LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 

OK 


ECLECTIC  EDUCATIONAL  SERIES. 


RAY'S 


NEW  HIGHER 


ARITHMETIC 


A  EEVISED  EDITION  OF  THE  HIGHER  ARITHMETIC 


BY 


JOSEPH  RAY,  M.  D. 

Late  Professor  in   Woodward  College. 


VAN  ANTWERP,  BRAGG  &  CO. 

CINCINNATI  AND  NEW  YORK. 


RAY'S  MATHEMATICAL  SERIES. 


ARITHMETIC. 

Ray's  New  Primary  Arithmetic. 
Ray's  New  Intellectual  Arithmetic. 
Ray's  New  Practical  Arithmetic. 
Ray's  New  Higher  Arithmetic. 

TWO-BOOK    SERIES. 

Ray's  New  Elementary  Arithmetic. 
Ray's  New  Practical  Arithmetic. 


ALGEBRA. 

Ray's  New  Elementary  Algebra. 
Ray's  New  Higher  Algebra. 


HIGHER   MATHEMATICS 

Ray's  Plane  and  Solid  Geometry. 

Ray's  Geometry  and  Trigonometry. 

Ray's  Analytic  Geometry. 

Ray's  Elements  of  Astronomy. 

Ray's  Surveying  and  Navigation. 

Ray's  Differential  and  Integral  Calculus. 


COPYRIGHT, 

1880, 
BY  VAN  ANTWERP,  BRAGG  &  Co. 


PREFACE. 


KAY'S  HIGHER  ARITHMETIC  was  published  nearly  twenty-five 
years  ago.  Since  its  publication  it  has  had  a  more  extensive  circu- 
lation than  any  other  similar  treatise  issued  in  this  country.  To 
adapt  it  more  perfectly  to  the  wants  of  the  present  and  future,  it  has 
been  carefully  revised. 

It  has  been  the  aim  of  the  revision  to  make  BAY'S  NEW  HIGHER 
ARITHMETIC  thoroughly  practical,  useful,  and  teachable.  To  this 
end  the  greatest  care  has  been  given  to  securing  concise  definitions 
and  explanations,  and,  at  the  same  time,  the  systematic  and  thorough 
presentation  of  each  subject.  The  pupil  is  taught  to  think  for  him- 
self correctly,  and  to  attain  his  results  by  the  shortest  and  best 
methods.  Special  attention  is  given  to  modern  business  transactions, 
and  all  obsolete  matter  has  been  discarded. 

Almost  every  chapter  of  the  book  has  been  entirely  rewritten, 
without  materially  changing  the  general  plan  of  the  former  edition, 
although  much  new,  and  some  original  matter  has  been  introduced. 
Many  of  the  original  exercises  are  retained. 

Particular  attention  is  called  to  the  rational  treatment  of  the 
Arithmetical  Signs,  to  the  prominence  given  to  the  Metric  System, 
and  to  the  comprehensive,  yet  practical,  presentation  of  Percentage 
and  its  various  Applications.  The  method  of  combining  the  algebraic 
and  geometric  processes  in  explaining  square  and  cube  root  will  com- 
mend itself  to  teachers.  The  chapter  on  Mensuration  is  unusually 

full  and  varied,  and  contains  a  vast  amount  of  useful  information. 

(iii) 

111908 


IV 


PREFACE. 


The  Topical  Outlines  for  Review  will  prove  invaluable  to  both 
teachers  and  pupils  in  aiding  them  to  analyze  and  to  classify  their 
arithmetical  knowledge  and  to  put  it  together  so  as  to  gain  a  com- 
prehensive view  of  it  as  a  whole. 

Principles  and  Formulas  are  copiously  interspersed  as  summaries, 
to  enable  pupils  to  work  intelligently. 

The  work,  owing  to  its  practical  character,  logical  exactness,  and 
condensation  of  matter,  will  be  found  peculiarly  adapted  to  the  wants 
of  classes  in  High  Schools,  Academies,  Normal  Schools,  Commercial 
Schools  and  Colleges,  as  well  as  to  private  students. 

The  publishers  take  this  opportunity  of  expressing  their  obligations 
to  J.  M.  GREENWOOD,  A.  M.,  Superintendent  of  Public  Schools,  Kan- 
sas City,  Mo.,  who  had  the  work  of  revision  in  charge,  and  also  to 
REV.  DR.  U.  JESSE  KNISEL.Y,  of  Newcomerstown,  Ohio,  for  his 
valuable  assistance  in  revising  the  final  proof-sheets. 

CINCINNATI,  July,  1880. 


CONTENTS. 


PAGE 

I.  INTRODUCTION 9 

II.  NUMERATION  AND  NOTATION 15 

III.  ADDITION 23 

IV.  SUBTRACTION  . .27 

Business  Terms  and  Explanations 29 

V.  MULTIPLICATION .        .        .      31 

When  the  multiplier  does  not  exceed  12  .  .  .  .32 
When  the  multiplier  exceeds  12 .  .  .  .  .34 
Business  Terms  and  Explanations  .  .  .  .  .36 
Contractions  in  Multiplication 38 

VI.  DIVISION 43 

Long  Division 45 

Short  Division 47 

Contractions  in  Division 49 

Arithmetical  Signs 50 

General  Principles       ........       52 

Contractions  in  Multiplication  and  Division       .        .        .53 

VII.  PROPERTIES  OF  NUMBERS     .        .        .        .        .        .        .59 

Factoring 61 

Greatest  Common  Divisor 64 

Least  Common  Multiple ".68 

Some  Properties  of  the  Number  Nine  ^  ...  70 
Cancellation 72 

VIII.  COMMON  FRACTIONS 75 

Numeration  and  Notation  of  Fractions       .         .         .         .77 

Keduction  of  Fractions 78 

Common  Denominator 82 

Addition  of  Fractions 85 

Subtraction  of  Fractions 86 

Multiplication  of  Fractions 87 

(v) 


vi  CONTENTS. 

PAGE 

Division  of  Fractions 90 

The  G.  C.  D.  of  Fractions .        .92 

The  L.  C.  M.  of  Fractions    .        .        .        .        .        .        .94 

IX.  DECIMAL  FRACTIONS      . 99 

Numeration  and  Notation  of  Decimals       .         .         .         .-  100 

Eeduction  of  Decimals        . 103 

Addition  of  Decimals 105 

Subtraction  of  Decimals      .        .        .        .        .         .        .  106 

Multiplication  of  Decimals         .        .        .        .        .        .  108 

Division  of  Decimals Ill 

X.  CIRCULATING  DECIMALS 115 

Keduction  of  Circulates 118 

Addition  of  Circulates         .        .        .        .        .        .        .120 

Subtraction  of  Circulates 121 

Multiplication  of  Circulates        .        .        .         .        .     >   .  122 

Division  of  Circulates 123 

XI.  COMPOUND  DENOMINATE  NUMBERS       .        .        .        .        .  125 

Measures  of  Value 125 

Measures  of  Weight .        .  130 

Measures  of  Extension 133 

Measures  of  Capacity 139 

Angular  Measure 142 

Measure  of  Time 143 

Comparison  of  Time  and  Longitude            ....  146 

Miscellaneous  Tables 146 

The  Metric  System       .        . 147 

Measure  of  Length .  149 

Measure  of  Surface 149 

Measure  of  Capacity 150 

Measure  of  Weight    .         .  * 150 

Table  of  Comparative  Values 152 

Reduction  of  Compound  Numbers 154 

Addition  of  Compound  Numbers         .....  160 

Subtraction  of  Compound  Numbers     .         .         .         .  163 

Multiplication  of  Compound  Numbers        ....  165 

Division  of  Compound  Numbers 167 

Longitude  and  Time    .         .         .         .         .         .         .         .  169 

Aliquot  Parts       .         . 172 


CONTENTS.  vii 


XII. 

RATIO  .        .        .        .        .        .                .        .        . 

PAGE 

175 

XIII 

.  PROPORTION        

.     177 

Simple  Proportion       

.     178 

Compound  Proportion          .         .         .         . 

.    184 

XIV. 

PERCENTAGE       

.     188 

Additional  Formulas           

.     197 

Applications  of  Percentage          .         . 

.    197 

XV. 

PERCENTAGE.  —  APPLICATIONS.     (  Without  Time.) 

.    199 

I.        Profit  and  Loss            ...... 

.    199 

ii.       Stocks  and  Bonds        

.     204 

m.     Premium  and  Discount      

.    208 

IV.      Commission  and  Brokerage        .... 

.     213 

V.       Stock  Investments       

.     220 

vi.      Insurance    ........ 

.     228 

vn.    Taxes          

.     232 

vni.   United  States  Revenue       

.    236 

XVI. 

PERCENTAGE.  —  APPLICATIONS.     (  With  Time.)    . 

.    242 

I.    Interest    .         

.     242 

Common  Method       

.     245 

Method  by  Aliquot  Parts         .... 

.    246 

Six  Per  Cent  Methods       

.    246 

Promissory  Notes      

.     254 

Annual  Interest        

.    259 

ii.    Partial  Payments           

.    261 

U.S.  Rule         

.     261 

Connecticut  Rule      

.    264 

Vermont  Rule            

.     265 

Mercantile  Rule        

.    266 

in.    True  Discount     

.    266 

iv.    Bank  Discount     

.     268 

V.      Exchange     

.    278 

Domestic  Exchange  ...... 

.    279 

Foreign  Exchange     

.     281 

Arbitration  of  Exchange  

.     283 

vr.    Equation  of  Payments          

.    286 

vir.  Settlement  of  Accounts         

.     292 

Account  Sales    . 

.     296 

297 

viii  CONTENTS. 

PAGE 

Viii.  Compound  Interest 298 

ix.      Annuities 308 

Contingent  Annuities       .         .         .         .         .         .  317 

Personal  Insurance   .......  322 

XVII.  PARTNERSHIP 327 

Bankruptcy 331 

XVIII.  ALLIGATION      .        -. 333 

Alligation  Medial 333 

Alligation  Alternate 334 

XIX.  INVOLUTION 342 

XX.  EVOLUTION 347 

Extraction  of  the  Square  Koot    .        .        .        .        .        .  349 

Extraction  of  the  Cube  Root 354 

Extraction  of  Any  Koot 359 

Horner's  Method 360 

Applications  of  Square  Koot  and  Cube  Koot      .         .         .  363 

Parallel  Lines  and  Similar  Figures 366 

XXI.  SERIES 369 

Arithmetical  Progression 369 

Geometrical  Progression      .         .                 .         .         .         .  373 

XXII.  MENSURATION .        .        .  378 

Lines 378 

Angles .        .        .379 

Surfaces 379 

Areas  .        .        .        .        .        .        .        .         .        .        .  382 

Solids 390 

Miscellaneous  Measurements 395 

Masons'  and  Bricklayers'  Work        ....  395 

Gauging .        .396 

Lumber  Measure 398 

To  Measure  Grain  or  Hay 398 

XXIII.  MISCELLANEOUS  EXERCISES    .                                        .  401 


RAT'S 
HIGHER    ARITHMETIC. 


I.  INTEODUOTIOK 

Article  1.  A  definition  is  a  concise  description  of  any 
object  of  thought,  and  must  be  of  such  a  nature  as  to  dis- 
tinguish the  object  described  from  all  other  objects. 

2.  Quantity  is   any  thing    which   can   be    increased  or 
diminished;    it  embraces  number  and  magnitude.     Number 
answers  the  question,    "How  many?"     Magnitude,   "How 
much?" 

3.  Science  is  knowledge  properly  classified. 

&.  The  primary  truths  of  a  science  are  called  Prin- 
ciples. 

5.  Art  is  the  practical  application  of  a  principle  or  the 
principles  of  science. 

6.  Mathematics  is  the  science  of  quantity. 

7.  The  elementary  branches  of   mathematics  are  Arith- 
metic, Algebra,  and  Geometry. 

8.  Arithmetic  is  the  introductory  branch  of  the  science 
of  numbers.     Arithmetic  as  a  science  is  composed  of  defini- 

(9) 


10  RAY'S  HIGHER  ARITHMETIC. 

tions,  principles,  and  processes  of  calculation ;  as  an  art,  it 
teaches  how  to  apply  numbers  to  theoretical  and  practical 
purposes. 

9.  A  Proposition  is  the  statement  of  a  principle,  or  of 
something  proposed  to  be  done. 

10.  Propositions   are    of  two    kinds,    demonstrable    and 
indemonstrable. 

Demonstrable  propositions  can  be  proved  by  the  aid  of 
reason.  Indemonstrable  propositions  can  not  be  made  simpler 
by  any  attempt  at  proof. 

11.  An  Axiom  is  a  self-evident  truth. 

12.  A  Theorem  is  a  truth  to  be  proved. 

13.  A  Problem  is  a  question  proposed  for  solution. 

14.  Axioms,  theorems,  and  problems  are  propositions. 

15.  A  process  of  reasoning,  proving  the  truth  of  a  prop- 
osition, is  called  a  Demonstration. 

• 

16.  A  Solution  of  a  problem  is  an  expressed  statement 

showing  how  the  result  is  obtained. 

17.  The  term  Operation,  as  used  in  this  book,  is  applied 
to  illustrations  of  solutions. 

18.  A  Rule  is  a  general  direction  for  solving  all  prob- 
lems of  a  particular  kind. 

19.  A  Formula  is  the  expression  of  a  general  rule  or 
principle  in  algebraic  language;  that  is,  by  symbols. 

20.  A  Unit  is  one  thing,  or  one.       One  thing  is  a  con- 
crete unit ;  one  is  an  abstract  unit. 

21.  Number    is    the    expression  of   a  definite  quantity. 
Numbers  are  either  abstract  or  concrete.     An  abstract  num- 
ber is  one  in  which  the  kind  of  unit  is  not  named ;  a  concrete 
number  is  one  in  which  the  kind  of  unit  is  named.      Con- 
crete numbers  are  also  called  Denominate  Numbers. 


INTE  OD  UCTION.  11 

22.  Numbers  are  also  divided  into  Integral,  Fractional, 
and  Mixed. 

An  Integral  number,  or  Integer,  is  a  whole  number;  a 
Fractional  number  is  an  expression  for  one  or  more  of  the 
equal  parts  of  a  divided  whole;  a  Mixed  number  is  an 
Integer  and  Fraction  united. 

23.  A  Sign  is  a  character  used  to  show  a  relation  among 
numbers,  or  that  an  operation  is  to  be  performed. 

24.  The  signs  most  used  in  Arithmetic  are 

+                                      -r-          V 
=        :         ::         ()         '. 


25.  The  sign  of  Addition  is  [+],  and  is  called  plus.     The 
numbers    between    which    it    is    placed    are    to    be    added. 
Thus,  3  +  5  equals  8. 

Plus  is  described  as  a  perpendicular  cross,  in  which  the  bisecting 
lines  are  equal. 

26.  The  sign  of  Subtraction  is  [ — ],  and  is  called  minus. 
When  placed  between  two  numbers,  the  one  that  follows  it 
is  to  be  taken  from  the  one  that  precedes  it.     Thus,  7  —  4 
equals  3. 

Minus  is  described  as  a  short  horizontal  line. 

Plus  and  Minus  are  Latin  words.  Plus  means  more;  minus  means 
less. 

Michael  Steifel,  a  German  mathematician,  first  introduced  + 
and  —  in  a  work  published  in  1544. 

27.  The  sign  of  Multiplication  is  [  X  ]>  and  is  read  mul- 
tiplied %,  or  times.     Thus,  4  X  5  is  to  be  read,  4  multiplied 
bij  5,  or  4  times  5. 

The  sign  is  described  as  an  oblique  cross. 

William  Oughtred,  an  Englishman,  born  in  1573,  first  introduced 
the  sign  of  multiplication. 

28.  The  sign  of  Division  is  [-T-],  and  is  read  divided  by. 
When  placed  between  two  numbers,  the  one  on  the  left  is 


12  RA  Y'S  HIGHER  ARITHMETIC. 

to   be    divided   by  the   one    on    the    right.      Thus,  20-|-4 
equals  5. 

The  sign  is  described  as  a  short  horizontal  line  and  two  dots ; 
one  dot  directly  above  the  middle  of  the  line,  and  the  other  just 
beneath  the  middle  of  it. 

Dr.  John  Pell,  an  English  analyst,  born  in  1610,  introduced  the 
sign  of  division. 

29.  The  Radical  sign,  [  j/  ]»  indicates  that  some  root  is 
to  be  found.     Thus,  ^/36  indicates  that  the  square  root  of 
36  is  required ;    ^125,  that    the    cube  root  of  125  is  to  be 
found;    and  iX625  indicates  that  the  fourth  root  of  625  is 
to  be  extracted. 

The  root  to  be  found  is  shown  by  the  small  figure  placed  between 
the  branches  of  the  Radical  sign.  The  figure  is  called  the  index. 

30.  The   signs,    +,    — ,    X,    -f-,    i/,    are   symbols   of 
operation. 

31.  The   sign  of  Equality  is  [=],  two  short    horizontal 
parallel  lines,   and    is   read    equals  or  is  equal  to,  and  sig- 
nifies  that  the  quantities  between   which   it  is   placed  are 
equal.     Thus,   3  +  5=9  —  1.     This  is  called  an    equation, 
because  the  quantity  3  +  5  is  equal  to  9  —  1. 

32.  Ratio  is  the  relation  which  one  number  bears  to 
another  of  the  same  kind.    The  sign  of  Ratio  is  [:  ].     Ratio 
is  expressed  thus,  6  :  3  =  f  =  2,  and  is  read,   the  ratio   of 
6  to  3=2,  or  is  2. 

The  sign  of  ratio  may  be  described  as  the  sign  of  division  with 
the  line  omitted.  It  has  the  same  force  as  the  sign  of  division,  and 
is  used  in  place  of  it  by  the  French. 

33.  Proportion  is  an  equality  of  ratios.      The  sign  of 
Proportion   is    [:  :],    and    is    used   thus,    3  :  6  :  :  4  :  8;    this 
may  be  read,    3  is  to  6  as  4  is  to  8;  another  reading,  the 
ratio  of  3  to  6  is  equal  to  the  ratio  of  4  to  8. 


INTRODUCTION. 


13 


34.  The  signs  [(),  -     -],  are  signs  of  Aggregation — the 
first  is  the  Parenthesis,  the  second  the  Vinculum.     They  are 
used   for  the   same  purpose;    thus,  24 — (8  +  7),  or   24- 
8  +  7,   means   that  the    sum  of  8  +  7   is  to  be  subtracted 
from  24.     The  numbers  within    the   parenthesis,  or   under 
the  vinculum,  are  considered  as  one  quantity. 

35.  The  dots  [.  .  .  .],  used  to  guide  the  eye  from  words 
at  the  left  to  the  right,  are   called  Leaders,  or  the  sign  of 
Continuation,  and  are  read,  and  so  on. 

36.  The  sign  of  Deduction  is  [•'.],  and  is  read  therefore, 
hence,  or  consequently. 

37.  The   signs,  =,     :,     : :,     (  ), ,     .  .  .  .,     .-., 

are  symbols  of  relation. 

38.  Arithmetic  depends  upon  this  primary  proposition: 
that  any  number   may  be  increased  or   diminished.     "  In- 
creased" comprehends   Addition,    Multiplication,  and   Invo- 
lution; " decreased,"  Subtraction,   Division,  and   Evolution. 

39.  The   fundamental   operations  of  Arithmetic  in    the 
order  of  their  arrangement,  are :  Numeration  and  Notation, 
Addition,  Subtraction,  Multiplication,  and  Division. 


Topical  Outline. 
INTRODUCTION. 


1.  Definition. 

2.  Quantity. 

3.  Science. 

4.  Principles. 

5.  Art. 

6.  Mathematics. 

7.  Proposition. 

8.  Demonstration. 

9.  Solution. 


10.  Operation. 

11.  Rule. 

12.  Formula. 

13.  Unit. 

14.  Number. 

15.  Sign.       . 

16.  Signs  most  used. 

17.  Primary  Proposition. 

18.  Fundamental  Operations. 


14 


MAY'S  HIGHER  ARITHMETIC. 


Topical  Outline  of  Arithmetic, 


Preliminary  Definitions.. 


1.  Definition. 

2.  Quantity. 

3.  Science. 

4.  Mathematics. 

5.  Proposition. 

6.  Theorem. 

7.  Axiom. 

8.  Demonstration. 

9.  Solution. 

10.  Rule. 

11.  Sign. 


1.  As  a  Science. ..  J 


1.  Definitions. 


2.  Classification  of  Numbers , 

f  Numeration 

1.  J       and 

I  Notation. 

2.  Addition. 

3.  Subtraction. 

4.  Multiplication. 

5.  Division. 

6.  Involution. 

7.  Evolution. 


L  3.  Operations.    ^ 


f  Abstract. 

\  Concrete. 

r  Integral. 
2. 1  Fractional. 

(_  Mixed. 
3    f  Simple. 

I  Compound. 


2.  As  an  Art 


1.  Terms  often  used... 


2.  Signs. 


3.  Applications.. 


1.  Problem. 

2.  Operation 

3.  Solution. 

4.  Principle. 

5.  Formula. 

6.  Rule. 

7.  Proof. 


1.  To  Integers. 
2/  To  Fractions. 

3.  To  Compound  Numbers. 

4.  To  Ratio  and  Proportion. 

5.  To  Percentage. 

6.  To  Alligation. 

7.  To  Progression. 

8.  To  Involution  and  Evolution. 
^  9.  To  Mensuration. 


II.  NUMEKATKOT  AKD  NOTATION. 

40.  Numeration  is  the  method  of  reading  numbers. 

Notation  is  the  method  of  writing  numbers.  Numbers 
are  expressed  in  three  ways ;  viz. ,  by  words,  letters,  and  figures. 

41.  The  first  nine  numbers  are  each  represented  by  a 
single  figure,  thus : 

123456789 

one.      two.     three,    four.      five.      six.      seven,    eight,   nine. 

All  other  numbers  are  represented  by  combinations  of 
these  and  another  figure,  0,  called  zero,  naught,  or  cipher. 

REMARK. — The  cipher,  0,  is  used  to  indicate  no  value.  The  other 
figures  are  called  significant  figures,  because  they  indicate  some  value. 

42.  The  number  next  higher  than  9  is  named  ten,  and 
is  written  with  two  figures,  thus,  10 :  in  which  the  cipher,  0, 
merely  serves  to  show  that  the  unit,  1,  on  its  left,  is  different 
from  the  unit,  1,  standing  alone,  which  represents  a  single 
thing,  while  this,  10,  represents  a  single  group  of  ten  things. 

The  nine  numbers  succeeding  ten  are  written  and  named 
as  follows : 

11  12  13  14  15  16 

eleven,      twelve.       thirteen,      fourteen.      fifteen.        sixteen. 

17  18  19 

seventeen,     eighteen.      nineteen. 

In  each  of  these,  the  1  on  the  left  represents  a  group  of  ten 
things,  while  the  figure  on  the  right  expresses  the  units  or 
single  things  additional,  required  to  make  up  the  number. 

REMARK. — The  words  eleven  and  twelve  are  supposed  to  be  derived 
from  the  Saxon,  meaning  one  left  after  ten,  and  two  left  after  ten.  The 
words  thirteen,  fourteen,  etc.,  are  contractions  of  three  and  ten,  four  and 

ten,  etc. 

(15) 


16  RAY'S  HIGHER   ARITHMETIC. 

The  next  number  above  nineteen  (nine  and  ten),  is  ten 
and  ten,  or  two  groups  of  ten,  written  20,  arid  called  twenty. 

The  next  numbers  are  twenty-one,  21;  twenty-two,  22;  etc., 
up  to  three  tens,  or  thirty,  30 ;  forty,  40 ;  fifty,  50 ;  sixty,  60 ; 
seventy,  70 ;  eighty,  80 ;  ninety,  90. 

The  highest  number  that  can  be  written  with  two  figures 
is  99,  called  ninety-nine;  that  is,  nine  tens  and  nine  units. 

The  next  higher  number  is  9  tens  and  ten,  or  ten  tens, 
which  is  called  one  hundred,  and  written  with  three  figures, 
100;  in  which  the  two  ciphers  merely  show  that  the  unit  on 
their  left  is  neither  a  single  thing,  1,  nor  a  group  of  ten 
things,  10,  but  a  group  of  ten  tens,  being  a  unit  of  a  higher 
order  than  either  of  those  already  known. 

In  like  manner,  200,  300,  etc.,  express  two  hundreds,  three 
hundreds,  and  so  on,  up  to  ten  hundreds,  called  a  thousand, 
and  written  with  four  figures,  1000,  being  a  unit  of  a  still 
higher  order. 

43.  The  Order  of  a  figure  is  the  place  it  occupies  in  a 
number. 

From  what  has  been  said,  it  is  clear  that  a  figure 
in  the  1st  place,  with  no  others  to  the  right  of  it,  expresses 
units  or  single  things;  but  standing  on  the  left  of  another 
figure,  that  is,  in  the  2d  place,  expresses  groups  of  tens; 
and  standing  at  the  left  of  two  figures,  or  in  the  3d  place, 
expresses  tens  of  tens,  or  hundreds;  and  in  the  4th  place, 
expresses  tens  of  hundreds  or  thousands.  Hence,  counting 
from  the  right  hand, 

The  order  of  Units  is  in  the  1st  place,  1 

The  order  of  Tens  is  in  the  2d    place,  10 

The  order  of  Hundreds  is  in  the  3d    place,  100 

The  order  of  Thousands  is  in  the  4th  place,  1000 

By  this  arrangement,  the  same  figure  has  different  values 
according  to  ilie  place,  or  order,  in  which  it  stands.  Thus,  3 
in  the  first  place  is  3  units;  in  the  second  place  3  tens,  or 
thirty;  in  the  third  place  3  hundreds;  and  so  on. 


NUMERA  TION  AND  NO  TA  TION.  1 7 

44.  The  word  Unite  may  be  used  in  naming  all  the  orders, 
as  follows: 

Simple  units  are  called  Units  of  the  1st  order. 

Tens  "         "  Units  of  the  2d    order. 

Hundreds  "         "  Units  of  the  3d    order. 

Thousands  "         "  Units  of  the  4th  order. 
etc.                                              etc. 

45.  The   following  table  shows  the   place  and  name  of 
each  order  up  to  the  fifteenth. 

TABLE  OP  ORDERS. 

15th.  14th.  13th.  12th.  llth.  10th.  9th.  8th.  7th.  6th.  5th.  4th.  3d.  2d.  1st. 


cc 

a 

o 

H 

o 

CO 

Trillions  .  .  . 

• 

s  of  Billions.  . 

CO 

3 

s 

CO 

O 

O 

CO 

cc 

a 

o 

s 

s  of  Thousands. 

Thousands  .  . 

CO 

CO 

• 

• 

CO 

•a 

CO 

PJ 

2 

H3 

«•*-! 
O 

CO 

p 

o 

o> 
t4 
T3 
0 

*0 

co 

co 
R 

O 

S 

T3 

0          0 

%               ^H 

o> 

?H 

TS 

c 

o 

CO 

ed 

CO 

p 

Q 

0) 

n 

CO 

02 

S3 

B 

1 

g 

S 

B 

'I 

S 

B 

H      S 

2 

B 

0 

H 

B 

1 

P 

46.  For  convenience  in  reading  and  writing  numbers, 
orders  are  divided  into  groups  of  three  each,  and  each 
group  is  called  a  period.  The  following  table  shows  the 
grouping  of  'the  first  fifteen  orders  into  five  periods : 

TABLE  OF  PERIODS. 


o 


o;—;       o  ^H 


e^  .0  e^  .2  0 


08 


O 


«         S         H       °  p 


BHP    K^Jp'  WHP    WHP    BHP 
654    321    987    654    321 

5th  Period.    4th  Period.  3d  Period.  2d  Period.  1st  Period. 
H.  A.  2. 


18  RAY'S  HIGHER  ARITHMETIC. 

47.  It  will  be  observed  that  each  period  is  composed  of 
units,  tens,  and  hundreds  of  the  same  denomination. 

48.  List  of  the  Periods,  according  to  the  common   or 
French  method  of  Numeration. 


First      Period,  Units. 

Second        "  Thousands. 

Third          "  Millions. 

Fourth        "  Billions. 

Fifth  "  Trillions. 


Sixth       Period,  Quadrillions. 
Seventh        "         Quintillions. 
Eighth         "         Sextillions. 
Ninth  "         Septillions. 

Tenth  "         Octillions. 


The  next  twelve  periods  are,  Nonillions,  Decillions,  Undecillions, 
Duodecillions,  Tredecillions,  Quatuordecillions,  Quindecillions, 
Sexdecillions,  Septendecillions,  Octodecillions,  Novendecillions, 
Vigintillions. 

PRINCIPLES. — 1.  Ten  units  of  any  order  always  make  one 
of  the  next  higher  order. 

2.  Removing  a  significant  figure  one  place  to  the  left  increases 
its  value  tenfold;  one  place  to  the  right ,  decreases  its  value  ten- 
fold. 

3.  Vacant  orders  in  a  number  are  filled  with  ciphers. 

PROBLEM. — Express  in  words  the  number  which  is  repre- 
sented by  608921045. 

SOLUTION. — The  number,  as  divided  into  periods,  is  608'921*045; 
and  is  read  six  hundred  and  eight  million  nine  hundred  and 
twenty-one  thousand  and  forty-five. 

Rule  for  Numeration. — 1.  Begin  at  the  right,  and  point 
the  number  into  periods  of  three  figures  each. 

2.  Commence  at  the  left,  and  read  in  succession  each  period 
with  its  name. 

REMARK. — Numbers  may  also  be  read  by  merely  naming  each 
figure  with  the  name  of  the  place  in  which  it  stands.  This  method, 
however,  is  rarely  used  except  in  teaching  beginners.  Thus,  the 
numbers  expressed  by  the  figures  205,  may  be  read  two  hundred  and 
five,  or  two  hundreds  no  tens  and  five  units. 


NUMERATION  AND  NOT  A  TION.  19 

EXAMPLES  IN  NUMERATION. 

7  4053  204026  4300201 

40  7009  500050  29347283 

85  12345  730003  45004024 

278  70500  1375482  343827544 

1345  165247  6030564  830070320 

832045682327825000000321 

8007006005004003002001000000 

60030020090080070050060030070 

504630209102J800v70324d703'250'207 

PROBLEM. — Express  in  figures  the  number  four  million 
.twenty  thousand  three  hundred  and  seven.  4020307. 

SOLUTION. — Write  4  in  millions  period ;  place  a  dot  after  it  to 
separate  it  from  the  next  period  :  then  write  20  in  thousands  period  ; 
place  another  dot :  then  write  307  in  units  period.  This  gives 
4*20*307.  As  there  are  but  two  places  in  the  thousands  period,  a 
cipher  must  be  put  before  20  to  complete  its  orders,  and  the  number 
correctly  written,  is  4020307. 

NOTE. — Every  period,  except  the  highest,  must  have  three  figures ; 
and  if  any  period  is  not  mentioned  in  the  given  number,  supply  its 
place  with  three  ciphers. 

Rule  for  Notation. — Begin  at  the  left,  and  write  each 
period  in  its  proper  place — -filling  the  vacant  orders  with  ciphers. 

PROOF. — Apply  to  the  number,  as  written,  the  rule  for 
Numeration,  and  see  if  it  agrees  with  the  number  given. 

EXAMPLES  IN  NOTATION. 


1.  Seventy-five. 

2.  One  hundred  and  thirty-four. 

3.  Two  hundred  and  four. 

4.  Three  hundred  and  seventy. 

5.  One    thousand   two    hundred 
and  thirty-four, 


6.  Nine  thousand  and  seven. 

7.  Forty  thousand  five  hundred 
and  sixty-three. 

8.  Ninety  thousand  and  nine. 

9.  Two  hundred  and  seven  thou- 
sand four  hundred  and  one. 


20 


RAY'S  HIGHER  ARITHMETIC, 


10.  Six  hundred  and  forty  thou- 
sand and  forty. 

11.  Seven  hundred  thousand  and 
seven. 

12.  One    million     four    hundred 
and  twenty-one  thousand  six 
hundred  and  eighty-five. 

13.  Seven  million  and  seventy. 

14.  Ten     million     one     hundred 
thousand  and  ten. 

15.  Sixty  million  seven  hundred 
and  five  thousand. 

16.  Eight    hundred     and     seven 
million   forty   thousand    and 
thirty-one. 

17.  Two  billion  and  twenty  mill- 


18.  Nineteen  quadrillion  twenty 
trillion      and     five    hundred 
billion. 

19.  Ten    quadrillion    four   hun- 
dred     and       three      trillion 
ninety  billion    and  six  hun- 
dred  million. 

20.  Eighty    octillion    sixty  sex- 
tillion     three     hundred    and 
twenty-five     quintillion    and 
thirty-three  billion. 

21.  Nine  hundred  decillion  sev- 
enty iionillion    six    octillion 
forty   septillion    fifty    quad- 
rillion    two     hundred     and 
four     trillion     ten     million 
forty  thousand    and   sixty. 


ENGLISH  METHOD  OF  NUMERATION. 

49.  In  the  English  Method  of  Numeration  six  figures 
make  a  period.  The  first  period  is  units,  the  second  millions, 
the  third  billions,  the  fourth  trillions,  etc. 

The  following  table  illustrates  this  method : 


o 

*r^ 

^  -^ 

*0 

u 
I—  i 

-  -s 

1 

M 

i 

\  

c    g 
P 

/  

^; 

/  — 

^ 

N 

rfi 

H 

F 

EH 

H 

«-M 

^ 

S     O 

S  H 

Thousands 

Hundreds 
Tens 
Units 

o  ^; 

CO     H 

TJ 

<D     «4-l 

*H         O 

T3 

c  .g 

t§l 

Thousands 
Hundreds 

CO 

P 

^ 

CO 

"3 
P 

Hundreds  of 

Tens  of  Th. 
Thousands 

Hundreds 
Tens 
Units 

1^ 

^H  0 

-g  g 

MHS 

Thousands 
Hundreds 
Tens 
Units 

4  3 

2 

109 

8  7 

6  5 

4 

3 

2 

1  0 

987 

6  5 

4321 

By  this  system  the  twelve  figures  at  the  right  are  read,  two 
hundred  and  ten  thousand  nine  hundred  and  eighty-seven 


NUMERATION  AND  NOTATION. 


21 


million  six  hundred  and  fifty-four  thousand  three  hundred 
and  twenty-one.  By  the  French  method  they  would  be  read, 
two  hundred  and  ten  billion  nine  hundred  and  eighty-seven 
million  six  hundred  and  fifty-four  thousand  three  hundred 
and  twenty-one. 

ROMAN  NOTATION. 

50.  In  the  Roman  Notation,  numbers  are  represented  by 
seven  letters.  The  letter  I  represents  one;  V,  five;  X,  ten; 
L,  fifty;  C,  one  hundred;  D,  five  hundred;  and  M,  one  thou- 
sand. The  other  numbers  are  represented  according  to  the 
following  principles  : 

1st.  Every  time  a  letter  is  repeated,  its  value  is  repeated. 
Thus,  II  denotes  two;  XX  denotes  twenty. 

2d.  Where  a  letter  of  less  value  is  placed  before  one  of 
greater  value,  the  less  is  taken  from  the  greater;  thus,  IV 
denotes  four. 

3d.  Where  a  letter  of  less  value  is  placed  after  one  of 
greater  value,  the  less  is  added  to  the  greater;  thus,  XI 
denotes  eleven. 

4th.  Where  a  letter  of  less  value  stands  between  two  letters 
of  greater  value,  it  is  taken  from  the  following  letter, 
not  added  to  the  preceding;  thus,  XIV  denotes  fourteen, 
not  sixteen. 

5th.  A  bar  [ — ]  placed  over  a  letter  increases  its  value  a 
thousand  times.  Thus,  V  denotes  five  thousand;  M  denotes 
one  million. 

ROMAN  TABLE. 


I  . One. 

II Two. 

Ill Three. 

IV Four. 

V Five. 

VI .     .  Six. 

IX Nine. 

X  ........  Ten. 


XI Eleven. 

XIV Fourteen. 

XV Fifteen. 

XVI Sixteen. 

XIX Nineteen. 

XX        Twenty. 

XXI Twenty-one. 

XXX     .....  Thirty. 


22 


RAY'S  HIGHER  ARITHMETIC. 


XL 
L 

LX 
XC 

C 

cccc 

D 


ROMAN  TABLE.     (CONTINUED.) 


JL   \JJi  \jj  , 

Fifty. 

DCC      .     . 

.     .     Seven  hundred. 

Sixty. 

DCCC  .     . 

.     .     Eight  hundred. 

Ninety. 

DCCCC     . 

.     .     Nine  hundred. 

One  hundred. 

M          .    . 

.     One  thousand. 

Four  hundred. 

MM      .     . 

.     .     Two  thousand. 

Five  hundred.     MDCCCLXXXI        1881. 


1.  Definitions. 


Topical   Outline. 
NUMERATION  AND  NOTATION. 


-  1.  Arabic 


2.  Methods.  - 


2.  Roman. 


I 


1.  Characters 


2.  Terms 

3.  Principles. 

4.  Rules. 
1.  Names. 


i.  Names. 


2.  Values. 


{i 


Order. 
Periods. 


1.  Significant. 

2.  Zero. 

1.  Simple. 

2.  Local. 


{J: 


French. 
English. 


1.  Alone. 

2.  Repeated. 

3.  Preceding. 

4.  Following. 

5.  Between. 

6.  Line  Above. 


3.  Ordinary  Language. 


HI.  ADDITION. 

51.  Addition    is   the    process    of   uniting   two  or  more 
like  numbers  into  one  equivalent  number. 

Sum  or  Amount  is  the  result  of  Addition. 

52.  Since  a  number  is  a  collection  of  units  of  the  same 
kind,   two  or  more  numbers  can  be  united  into  one   sum,  only 
when  their  units  are  of  the   same   kind.       Two   apples   and  3 
apples  are  5  apples;  but  2  apples  and  3  peaches  can  not  be 
united  into  one  number,  either  of  apples  or  of  peaches. 

Nevertheless,  numbers  of  different  names  may  be  added  together, 
if  they  can  be  brought  under  a  common  denomination. 

PRINCIPLES. — 1.    Only  like  numbers  can  be  added. 

2.  The  sum  is  equal  to  all  the  units  of  all  the  parts. 

3.  TJie  sum  is  the  same  in  kind  as  the  numbers  added. 

4.  Units   of  the  same   order,  and  only  such,  can  be  added 
directly. 

5.  TJie   sum  is  the  same   in  whatever  succession  the  numbers 
are  added. 

REMARK. — Like  numbers,  similar  numbers,  and  numbers  of  the  same 
kind  are  those  having  the  same  unit. 

PROBLEM. —What  is  the  sum  of  639,  82,  and  543? 

SOLUTION. — Writing  the  numbers  as  in   the  margin,  OPERATION. 

say,  3,  5,  14  units,  which  are  1  ten  and  4  units.     Write  639 

the  4  units  beneath,  and  carry  the  1  ten  to  the  next  8  2 

column.     Then  1,.5,  13,  16  tens,  which  are  6  tens  to  be  543 

written  beneath,  and   1  hundred    to    be  carried    to    the  1264 
next    column.      Lastly,   1,  6,  12    hundreds,  which   is  set 
beneath,  there  being  no  other  columns  to  carry  to  or  add. 

DEMONSTRATION. — 1.  Figures  of  the  same  order  are  written  in  the 

(23) 


24  RAY'S  HIGHER  ARITHMETIC. 

same  column  for  convenience,  since  none  but  units  of  the  same  name 
can  be  added.     (Art.  52.) 

2.  Commence  at  the  right  to  add,  so  that  when  the  sum  of  any 
column  is  greater  than  nine,  the  tens  may  be   carried  to  the  next 
column,  and,  thereby,  units  of  the  same  name  added  together. 

3.  Carry  one  for  every  ten,  since  ten  units  of  each  order  make  one 
unit  of  the  order  next  higher. 

Rule  for  Adding  Simple  Numbers. — 1.  Write  the  num- 
bers to  be  added,  so  that  figures  of  the  same  order  may  stand 
in  the  same  column,  and  draw  a  line  directly  beneath. 

2.  Begin  at  the  right  and  add  each  column  separately, 
writing  the  units  wider  the  column  added,  and  carrying  the 
tens,  if  any,  to  the  next  column.  At  the  last  column  write  the 
last  whole  amount. 

METHODS  OF  PROOF. — 1.  Add  the  figures  downward 
instead  of  upward;  or 

2.  Separate    the    numbers    into    two    or  more    divisions ; 
find    the  sum   of  the   numbers  in  each  division,  and  then 
add  the  several  sums  together;  or 

3.  Commence  at  the  left;  add  each   column   separately; 
place  each  sum  under  that  previously  obtained,  but  extend- 
ing one   figure   further  to   the  right,  and   then  add  them 
together. 

In  each  of  these  methods  the  result  should  be  the  same 
as  when  the  numbers  are  added  upward. 

NOTE. — For  proof  by  casting  out  the  9's,  see  Art.  105. 

EXAMPLES  FOR  PRACTICE. 
Find  the  sum, 

1.  Of  76767;  7654;  50121;  775.  Ans.  135317. 

2.  Of  97674;  686;  7676;  9017.  Ans.  115053. 

3.  Of  971;  7430;  97476;  76734.  Ans.  182611. 

4.  Of  999;  3400;  73;  47;  452;  11000;  193;  97;  9903; 
42 ;  and  5100.  Ans.  31306. 


ADDITION.  25 

5.  Of  four  hundred  and  three ;  5025 ;  sixty  thousand  and 
seven ;    eighty-seven  thousand ;    two   thousand   and  ninety ; 
and  100.  154625. 

6.  Of  20050 ;    three  hundred   and  seventy  thousand  two 
hundred ;  four  million  and  five ;  two  million  ninety  thousand 
seven    hundred    and    eighty;    one    hundred    thousand    and 
seventy ;  98002 ;  seven  million  five  thousand  and  one ;  and 
70070.  13754178. 

7.  Of  609505 ;  90070  ;  90300420 ;  9890655  ;  789 ;  37599 ; 
19962401;  5278;  2109350;  41236;  722;  8764;  29753;  and 
370247.  123456789. 

8.  Of  two  hundred  thousand  two  hundred ;  three  hundred 
million  six   thousand  and   thirty ;  seventy  million  seventy 
thousand  and  seventy ;  nine  hundred  and  four  million  nine 
thousand  and  forty;  eighty  thousand;    ninety   million  nine 
thousand;    six  hundred  thousand  and  sixty;   five  thousand 
seven   hundred.  1364980100. 

In  each  of  the  7  following  examples,  find  the  sum  of  the 
consecutive  numbers  from  A  to  B,  including  these  numbers : 

A.  B. 

9.         119  131.  Ans.        1625. 

10.  987  1001.  Ans.       14910. 

11.  3267  3281.  Ans.  49110. 

12.  4197  4211.  Ans.  63060. 

13.  5397  5416.  Am.  108130. 

14.  7815  7831.  Ans.  132991. 

15.  31989  32028.  Ans.  1280340. 

16.  Paid  for  coffee,  $375;  for  tea,  $280;  for  sugar,  $564; 
for  molasses,  $119;  and  for  spices,  $75:  what  did  the  whole 
amount  to?  $1413. 

17.  I  bought  three  pieces  of  cloth:  the  first  cost  $87;  the 
second,  $25  more  than  the  first;    and  the  third,  $47  more 
than  the  second:  what  did  all  cost?  $358. 

18.  A  man  bought  three  bales  of  cotton.     The  first  cost 

H.  A.  3. 


26  -B^  Y'S  HIGHER  ARITHMETIC 

$325;  the  second  cost  $16  more  than  the  first;  and  the 
third,  as  much  as  both  the  others:  what  sum  was  paid  for 
the  three  bales?  $1332. 

19.  A  has  $75;  B  has  $19  more  than  A;  C  has  as 
much  as  A  and  B,  and  $23  more;  and  D  has  as  much  as 
A,  B,  and  C  together:  what  sum  do  they  all  possess? 

$722. 
. 

SUGGESTIONS. — Two  things  are  of  the  greatest  importance  in 
arithmetical  operations, — absolute  accuracy  and  rapidity.  The 
figures  should  always  be  plain  and  legible.  Frequent  exercises  in 
adding  long  columns  of  figures  are  recommended ;  also,  practice  in 
grouping  numbers  at  sight  into  tens  and  twenties  is  a  useful  exercise. 

Accountants  usually  resort  to  artifices  in  Addition  to  save  time 
and  extra  labor,  such  as  writing  the  number  to  be  carried  in  small 
figures  beneath  the  column  to  which  it  belongs,  also  writing  the 
whole  amount  of  each  column  separately,  etc. 


Topical  Outline. 
ADDITION. 

1.  Definitions. 

2.  Sign. 

3.  Principles. 

f  1.  Writing  the  Numbers. 

4.  Operation..-^  2.  Drawing  Line  Beneath. 

I  3.  Adding,  Reducing,  etc. 

5.  Rule. 

6.  Methods  of  Proof. 


IV.  SUBTKACTION. 

53.  Subtraction  is  the  process  of  finding  the  difference 
between  two  numbers  of  the  same  kind. 

The  larger  number  is  the  Minuend;  the  less,  the  Subtra- 
Jiend;  the  number  left,  the  Difference  or  Remainder. 

Minuend  means  to  be  diminished;  subtrahend,  to  be  sub- 
tracted. 

54.  Subtraction    is  the  reverse   of  Addition,  and  since 
none  but  numbers  of  the  same  kind  can  be  added  together 
(Art.  52),  it  follows,  therefore,  that  a  number  can  be  sub- 
tracted only  from  another  of  the  same  kind :    2  cents  can 
not  be  taken  from  5  apples,  nor  3  cows  from  8  horses. 

PRINCIPLES. — 1.  The  minuend  and  subtrahend  must  be  of  the 
same  kind. 

2.  The  difference  is    the  same  in  kind  as  the  minuend  and 
subtrahend. 

3.  The  difference  equals  the  minuend   minus  the  subtrahend. 

4.  TJie  minuend  equals  the  difference  plus  the  subtrahend. 

5.  The  subtrahend  equals  the  minuend  minus  the  difference. 

PROBLEM. — From  827  dollars  take  534  dollars. 

OPERATION. 

SOLUTION. — After   writing    figures    of    the   same     $  8  2  7 
order  in  the  same  column,  say,  4  units  from  7  units        534 
leave  3  units.     Then,  as  3  tens  can  not  be  taken     $293  Rem. 
from  2  tens,  add  10  tens  to  the  2  tens,  which  make 
12  tens,  and  3  tens  from  12  tens  leave  9  tens.     To  compensate  for 
the  10  tens  added  to  the  2  tens,  add  one  hundred  (10  tens)  to  the  5 
hundreds,  and  say,  6  hundreds  from  8  hundreds  leaves  2  hundreds; 
and  the  whole  remainder  is  2  hundreds  9  tens  and  3  units,  or  293. 

DEMONSTRATION. — 1.  Since  a  number  can  be  subtracted  only  from 

(27) 


28  RAY'S  HIGHER  ARITHMETIC. 

another  of  the  same  kind  (Art.  54),  figures  of  the  same  name  are 
placed  in  the  same  column  to  be  convenient  to  each  other,  the 
less  number  being  placed  below  as  a  matter  of  custom. 

2.  Commence  at  the  right  to  subtract,,  so  that  if  any  figure  is 
greater  than  the  one  above  it,  the  upper  may  be  increased  by  10,, 
and  the  next  in  the  subtrahend  increased  by  1,  or,  as  some  prefer,  the 
next  in  the  minuend  decreased  by  1. 

Rule  for  Subtracting  Simple  Numbers. — 1.  Write  the 
less  number  under  the  greater,  placing  units  under  units,  tens 
under  tens,  etc.,  and  draw  a  line  directly  beneath. 

2.  Begin  at  the  right,  subtract  each  figure  from  the  one  above 
it,  placing  the  remainder  beneath. 

3.  If  any  figure   exceeds    the   one   above  it,   add  ten  to  the 
upper,  subtract  the  lower  from  the  sum,  increase  by  1  the  units 
of  the  next  order  in  the  subtrahend,  and  proceed  as  before. 

PROOF. — Add  the  remainder  to  the  less  number.  If  the 
work  be  correct,  the  sum  will  be  equal  to  the  greater. 

NOTE. — For  proof  by  casting  out  the  9's,  see  Art.  105. 

EXAMPLES  FOR  PRACTICE. 

1.  From  30020037  take  50009. 

OPERATION. 

KEMARK.— When  there  are  no  figures  in  30020037 
the  lower  number  to  correspond  with  those  50009 

in    the    upper,   consider   the   vacant  places  2  9  9  70  0  2  8  -Kern, 
occupied  by  zeros. 

2.  From  79685  take  30253.  Ans.   49432. 

3.  From  1145906  take  39876.  Ans.  1106030. 

4.  From  2900000  take  777888.  •  Ans.  2122112. 

5.  From  71086540  take  64179730.  Ans.  6906810. 

6.  From  101067800  take  100259063.  Ans.  808737. 

7.  How   many  years  from  the  discovery  of  America    in 
1492,  to  the  Declaration  in  1776  ?  284  years. 

8.  A  farm  that  cost  $7253,  was  sold  at  a  loss  of  $395  ; 
for  how  much  was  it  sold  ?  $6858. 


SUBTRACTION.  29 

9.  The    difference  of    two   numbers  is    19034,    and  the 
greater  is  75421 :  what  is  the  less?  56387. 

10.  How  many  times  can  the  number  285  be  subtracted 
from  1425  ?  5  times. 

11.  Which  is  the  nearer  number  to  920736;  1816045  or 
25427?  Neither.     Why? 

12.  From  a   tract  of  land   containing  10000   acres,   the 
owner  sold  to  A  4750  acres;  and  to  B   875  acres  less  than 
A:  how  many  acres  had  he  left?  1375  acres. 

13.  A,  B,  C,  D,  are  4  places  in  order  in  a  straight  line. 
From  A  to  D  is  1463  miles;  from  A  to  C,  728  miles;  and 
from  B  to  D,  1317  miles.     How  far  is  it  from  A  to  B,  from 
B  to  C,  and  from  C  to  D  ? 

A  to  B,  146  miles ;  B  to  C,  582  miles ;  C  to  D,  735  miles. 

BUSINESS  TEEMS  AND  EXPLANATIONS. 

55.  Book-keeping   is  the  science  and  art  of  recording 
business  transactions. 

56.  Business  records  are  called  Accounts,  and  are  kept 
in  books  called   Account  Books.      The  books   mostly  used 
are  the  Day-book  and  Ledger.     In  the  Day-book  are  recorded 
the  daily  transactions  in  business,  and   the  Ledger  is  used 
to  classify  and  arrange  the  results  of  all  transactions  under 
distinct  heads. 

57.  Each  account  has  two  sides :  Dr. — Debits,  and  Cr. — 
Credits.      Sums  a  person  owes  are  his  Debits;    sums  owing 
to  him  are  his  Credits.      The  difference    between  the   sum 
of   the   Debits    and    the  sum  of   the   Credits,  is  called  the 
Balance.     Debits   are  preceded  by  "To,"  and  Credits   by 
"By." 

58.  Finding    the    difference    between    the    sum    of   the 
Debits  and    the  sum    of   the  Credits,  and  writing  it  under 
the  less  side  as  Balance,  is  called  Balancing. 


30 


KAY'S  HIGHER  ARITHMETIC. 


DR. 


PRACTICAL  EXERCISES. 

JAMES  CRAIG. 


Balance  the  following  account: 

DR.  THOMAS  BALDWIN. 


CR. 


1878. 

1878. 

July   4 

To  Merchandise  . 

560  50 

July  5 

By  Cash   

550  50 

„      6 

„   Interest  .... 

24  90 

„    11 

„  Bills  Payable  .    . 

890  70 

„    10 

„  Sundries     .    .    . 

870  60 

„    26 

,,  Sundries     .    .    . 

310  80 

,,    25 

„   Merchandise  .    . 

320  10 

„    31 

.,   Cash   

100  00 

„    31 

Ditto     .    .    . 

12540 

"  Balance  .    .    .    - 

4950 

1878. 

1901  50 

1901  50 

Aug.  1 

To  Balance  .... 

49  50 

CR. 


1879. 

1879. 

Jan.    3 
„    16 

To  Merchandise  .    . 
„  Sundries     .    .    . 

810  30 
580  20 

Jan.   7 

n    20 

By  Sundries     .    •    . 
,,  Cash    .    . 

1000  00 
300  00 

„    25 

,,  Cash   

381  25 

„    31 

Merchandise 

225  20 

„    31 

,,   Merchandise  .    . 

60  75 

,,  Balance  .... 
1 

1879. 

Feb.    1 

To  Balance  .... 

Topical  Outline. 
SUBTRACTION. 


1.  Definition.          (  1.  Minuend. 

2.  Terms \   2.  Subtrahend. 


3.  Sign. 

4.  Principles. 

5.  Operation 

6.  Rule. 

7.  Proof. 

8.  Applications. 


3.  Difference  or  Remainder. 

1.  Writing  the  Numbers. 

2.  Drawing  Line  Beneath. 

3.  Subtracting  and  Writing  Difference. 


Y.  MULTIPLICATION. 

59.  1.  Multiplication  is   taking  one  number  as  many 
times  as  there  are  units  in  another ;  or, 

2.  Multiplication   is  a  short  method  of  adding  numbers 
that  are  equal. 

60.  The  number  to  be  taken,  is  called  the  Multiplicand; 
the  other  number,  the  Multiplier;  and  the  result  obtained, 
the  Product.     The  Multiplicand  and  Multiplier  are  together 
called  Factors  (makers),  because  they  make  the  Product. 

PROBLEM.— How   many  trees  in  3  rows,  each  containing 
42  trees? 

SOLUTION. — Since  3  rows  contain  3  times  OPERATION. 

as  many  trees  as  one  row,  take  42  three  First  row,     42  tree? 

times.    This  may  be  done  by  writing  42  Second  row,  4  2  trees, 

three  times,  and  then  adding.    This  gives  Third  row,    42  trees. 
126  trees  for  the  whole  number  of  trees.  126  trees. 

Instead,"  however,    of    writing  42    three 

times,  write  it  once;  then  placing  under  it  42  trees, 

the  figure  3,  the  number  of  times  it  is  to  be  3 

taken,  say,  3  times  2  are  6,  and  3  times  4  126  trees, 

are  12.    This  process  is  Multiplication. 

PRINCIPLES. — 1.   The   multiplicand  may  be  either  concrete 
or  abstract. 

2.  The  multiplier  must  always  be  an  abstract  number. 

3.  The  product  is  the  same  in  kind  as  the  multiplicand. 

4.  The  product  is  the  same,  whicliever  factor  is  taken  as  the 
multiplier. 

5.  The  partial  products  are  the  same  in  kind  as  the  multi- 
plicand. 

6.  The  sum  of  the  partial  products  is  equal  to  the  total  product. 

(31) 


32 


RAY'S  HIGHER  ARITHMETIC. 


MULTIPLICATION  TABLE. 


1 

5 

4 

7) 

3 

4 

6 

6 

,  V 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

2 

~3 
4 

6 

8 
12 

10 

12 

14 

16 

18 

20 

22 

24 

26 

28 

30 

32 

34 

36 

38 

40 

9 

15 

18 

21 

24 

27 

30 

33 

36 

39 

42 

45 

48 

51 

54 

57 

60 

8 

12 

16 

20 

24 

28 

32 

36 

40 

44 

48 

52 

56 

60 

64 

68 

72 

76 

80 

5 
6 

7 
8 
9 
10 

11 
12 

10 
12 

15 
18 
21 

20 
24 

25 

30 

35 

40 

45 

50 

55 

60 

65 

70 

75 

80 

85 

90 

95 

100 

30 

36 

42 

48 

54 

60 

66 

72 

78 

84 

90 

96 

102 

108 

114 

120 

14 
16 
18 

28 
32 

35 

42 

49 

56 

63 

70 

77 

84 

3l 

98 

105 

112 

119 

126 

133 

140 

24 

27 

40 

48 

56 

64 

72 

80 

88 

96 

104 

112 

120 

128 

136 

144 

152 

160 

36 

45 

54 

63 

72 

81 

90 

99 

108 

117 

126 

135 

144 

153 

162 

171 

180 

29 
22 
24 

26 

30 
33 
36 
39 
42 

40 

50 

60 

70. 

80 

90 

100 

110 

120 

130 

140 

150 

160 

170 

180 

190 

200 

44 

48 

55 

66 

77 

88 

99 

110 

121 

132 

143 

154 

165 

176 

187 

198 

209 

220 

60 

72 

84 

96 

108 

120 

132 

144 

156 

168 

180 

192 

204 

216 

228 

240 

13 
14 
15 
16 
17 
18 
19 
20 

52 

65 

78 

91 

104 

117 

130 

143 

156 

169 

182 

195 

208 

221 

234 

247 

260 

28 

36 

70 

84 

98 

112 

126 

140 

154 

168 

182 

196 

210 

224 

238 

252 

266 

280 

30 
32 

45 

48 

60 
64 

75 

90 

105 

120 

135 

150 

165 

180 

195 

210 

225 

240 

255 

270 

285 

300 

80 

96 

112 

128 

144 

160 

176 

192 

208 

224 

240 

256 

272 

288 

304 

320 

34 

36 
38 
40 

51 
54 

68 
72 

85 

102 

119 

136 

153 

170 

187 

204 

221 

238 

255 

272 

289 

306 

323 

340 

90 

108 

126 

144 

162 

180 

198 

216 

234 

252 

270 

288 

306 

324 

342 

360 

57 

60 

76 

eSO 

95 

114 

133 

152 

171 

190 

209 

228 

247 

266 

285 

304 

323 

342 

361 

380 

100 

120 

140 

160 

180 

200 

220 

240 

260 

280 

300 

320 

340 

360 

380 

400 

61.  Multiplication  is  divided  into  two  cases : 

1.  When  the  multiplier  does  not  exceed  12. 

2.  When  the  multiplier  exceeds  12. 

CASE    I. 

62.  When  the  multiplier  does  not  exceed  12, 

PROBLEM. — At  the  rate  of  53  miles  an  hour,  how  far 
will  a  railroad  car  run  in  four  hours  ? 

SOLUTION. — Here  say,  4  times  3  (units)  are  12  OPERATION. 

(units) ;  write  the  2  in  units'  place,  and  carry  the  5  3  miles. 

1   (ten) ;  then,   4    times    5  are    20,  and    1    carried  4 

makes  21  (tens),  and  the  work  is  complete.  212  miles. 

DEMONSTRATION. — The  multiplier  being  written  under  the  mul- 
tiplicand for  convenience,  begin  with  units,  so  that  if  the  product 
should  contain  tens,  they  may  he  carried  to  the  tens  j  and  so  on  for 
each  successive  order. 


MULTIPLICATION.  33 

Since  every  figure  of  the  multiplicand  is  multiplied,  therefore, 
the  whole  multiplicand  is  multiplied. 

Rule. — 1.  Write  the  multiplicand,  and  place  the  multiplier 
under  it,  so  that  units  of  the  same  order  shall  stand  in  the  same 
column,  and  draw  a  line  beneatJi. 

2.  Begin  with  units;  multiply  each  figure  of  the  multiplicand 
by  the  multiplier,  carrying  as  in  Addition. 

PROOF. — Separate  the  multiplier  into  any  two  parts; 
multiply  by  these  separately.  The  sum  of  the  products 
must  be  equal  to  the  first  product. 

EXAMPLES  FOR  PRACTICE. 

1.  195X3.  Am.  585. 

2.  3823X4.  Ans.  15292. 

3.  8765  X  5.  Ans.  43825. 

4.  98374  X  6.  Am.  590244. 

5.  64382X7.  Ans.  450674. 

6.  58765X8.  Ans.  470120.' 

7.  837941  x  9.  Ans.  7541469. 

8.  645703  X  10.  Ans.  6457030. 

9.  407649X11.  Ans.  4484139. 

10.  If  4  men  can  perform  a  certain  piece  of  work  in  15 
days,  how  long  will  it  require  1  man? 

SOLUTION. — One  man  must  work  four  times  as  long  as  four  men. 
4  X  15  days  =  60  days. 

11.  How   many  pages  in  a   half-dozen   books,   each   con- 
taining 336  pages?  2016  pages. 

12.  How  far  can  an  ocean  steamer  travel  in  a  week,  at 
the  rate  of  245  miles  a  day?  1715  miles. 

13.  What    is    the    yearly   expense    of    a    cotton-mill,   if 
$32053  are  paid  out  every  month  ?  $384636. 

14.  A  receives   from  his   business  an   average   of  $45  a 
day.     He  pays  three  clerks  $3 ;  three,  $9 ;  and  three,  $12 
a  week ;  other  expenses  amount  to  $4  a  day ;  what  are  his 
profits  for  one  week?  $174. 


34  £A  Y>S  HIGHER  ARITHMETIC. 


CASE    II. 

63.     When  the  multiplier  exceeds  12. 
PROBLEM.— Multiply  246  by  235. 

SOLUTION. — First    multiply    by    5  OPERATION. 

(units),  and  place  the  first  figure  of  246 

the  product,  1230,  under  the  5  (units).  235 

Then  multiply  by  3  (tens),  and  place  1230  product  by        5 

the   first  figure  of  the  product,  738,  738     product  by      30 

under  the  3  (tens).     Lastly,  multiply  492        product  by  2  0  0 

by  2  (hundreds),  and  place  the  first  57810  product  by  235 
figure  of  the  product,  492,  under  the  2 
(hundreds).    Then  add  these  several  products  for  the  entire  product. 

DEMONSTRATION. — The  0  of  the  first  product,  1230,  is  units  (Art. 
62).  The  8  of  the  second  product,  738,  is  tens,  because  3  (tens)  times 
6  =  6  times  3  (tens)  =  18  (tens)  ;  giving  8  (tens)  to  be  written  in  the 
tens'  column.  The  2  of  the  third  product,  492,  is  hundreds,  because  2 
(hundreds)  times  6  =  6  times  2  (hundreds)  =  12  (hundreds),  giving 
2  (hundreds)  to  be  written  in  the  hundreds'  column.  The  right- 
hand  figure  of  each  product  being  in  its  proper  column,  the  other 
figures  will  fall  in  their  proper  columns;  and  each  line  being  the 
product  of  the  multiplicand  by  a  part  of  the  multiplier,  their  sum 
will  be  the  product  by  all  the  parts  or  the  whole  of  the  multiplier. 

Rule. — 1.  Write  the  multiplier  wider  the  multiplicand, 
placing  fijures  of  the  same  order  in  the  same  column,  and  draw 
a  line  beneath. 

2.  Multiply  each  figure  of  the  multiplicand  by  each  figure  of 
die  multiplier  successively;  first  by  the  units'  figure,  then  by  the 
tens9  figure,  etc.;  placing  the  right-hand  figure  of  each  product 
under  that  figure  of  the  multiplier  which  produces  it,  then  draw 
a  line  beneath. 

3.  Add  the  several  partial  products  together;  their  sum  will  be 
the  required  product. 

METHODS  OP  PROOF.—!.  Multiply  the  multiplier  by  the 
multiplicand;  this  product  must  be  the  same  as  the  first 
product. 


MULTIPLICATION.  35 

2.  The  same  as  when  the  multiplier  does  not  exceed  12. 
NOTE. — For  proof  by  casting  out  the  9's,  see  Art.  105. 

REMARK.— Although  it  is  custom-  OPERATION. 

ary  to   use  the  figures  of   the  multi- 
plier in  regular  order  beginning  with  235 
units,  it  will  give  the  same  product         738      product  by      30 
to  use  them  in  any  order,  observing      492        product  by  2  0  0 

that  the  right-hand  figure  of  each  partial    _     1230  product  by 5 

product  must  be  placed  under  the  figure      57810  product  by  2  3  5 
of  the  multiplier  which  produces  it. 

EXAMPLES  FOR  PRACTICE. 

1.  7198X216.  Ans.  1554768. 

2.  8862  X  189.  Ans.  1674918. 

3.  7575X7575.  Am.  57380625. 

4.  15607X3094.  Ans.  48288058. 

5.  93186X4455.  Am.  415143630. 

6.  135790X24680.  Am.  3351297200. 

7.  3523725  X  2583.  Am.  9101781675. 

8.  4687319  X  1987.  Am.  9313702853. 

9.  9264397  X  9584.  Am.  88789980848. 

10.  9507340X7071.  Am.  67226401140. 

11.  1644405  X  7749.  Am.  12742494345. 

12.  1389294X8900.  Ans.  12364716600. 

13.  2778588  X  9867.  Am.  27416327796. 

14.  204265X562402.  Am.  114879044530. 

PRACTICAL  PROBLEMS. 

1.  In   a   mile    are   63360  inches:    how   many  inches  are 
there  in  the  circumference  of  the  earth  at  the  equator  if 
the  distance  be  25000  miles  ?  1584000000  inches. 

2.  The  flow  of  the  Mississippi  at  Memphis  is  about  434000 
cubic  feet  a   second :  required  the  weight  of  water  passing 
that  point  in  one  day  of  86400  seconds,  if  a  cubic  foot  of 
water  weigh  62  pounds?  2324851200000  pounds. 


36  RAY'S  HIGHER  ARITHMETIC. 

3.  John  Sexton  sold  25625  bushels  of  wheat,  at  $1.20  a 
bushel,  and  received  in  payment  320  acres  of  land,  valued 
at  $50  an  acre ;  60   head  of  horses,  valued  at  $65  a  head ; 
10   town    lots,    worth    $150   each ;    and    the    remainder    in 
money:  how  much  money  did  he  receive?  $9350. 

4.  If  light   comes    from   the    sun   to    the    earth    in   495 
seconds,  what   is  the  distance  from  the  earth   to   the    sun, 
light  moving  192500  miles  a  second  ?  95287500  miles. 

5.  If  3702754400  cubic  feet  of  solid  matter  is  deposited 
in  the  Gulf  of  Mexico  by  the  Mississippi  every  year,  what 
is  the  deposit  for  6000  years?  22216526400000  cu.  ft. 

6.  The   area    of   Missouri  is   65350    square    miles :    how 
many  acres  are  there  in  the   State,  allowing   640  acres  to 
each  square  mile?  41824000  acres. 

7.  In  the  United  States,  at  the  close  of  1878,  there  were 
81841  miles  of  railroad :  if  the  average  cost  of  building  be 
$50000   a  mile,  what   has  been   the  total  cost   of  building 
the  railroads  in  this  country?  $4092050000. 

8.  The  number  of  pounds   of  tobacco   produced   in  this 
country  in   1870  was   260000000.      If  this  were  manufact- 
ured  into    plugs   one    inch  wide   and   six  inches  long,  and 
four    plugs  weigh  a  pound,    what  would    be   the  length  in 
inches  of  the  entire  crop  ?  6240000000  inches. 


BUSINESS  TERMS   AND  EXPLANATIONS. 

64.  A  Bill  is   an  account    of  goods   sold   or  delivered, 
services  rendered,  or  work  done.     Usually  the  price  or  value 
is  annexed  to  each  article,  and  the  date  of  purchase  given. 

It  is  customary  to  write  the  total  amount  off  to  the  right, 
and  not  directly  under  the  column  of  amounts  added. 

65.  A  Receipt  is  a  written    acknowledgment   of  pay- 
ment.     The    common   form    consists    in   signing   the    name 
after  the   words  " Received  Payment"  written   at   the  foot 
of  the  bill. 


M  UL  TIPLICATTON. 


37 


1.  Joseph  Allen  bought  of  Seth  Ward,  at  Springfield, 
111.,  Jan.  2,  1879,  30  barrels  of  flour,  at  $3.60  a  barrel; 
48  barrels  of  mess  pork,  at  $16.25  a  barrel;  16  boxes  of 
candles,  at  $3.50  a  box;  23  barrels  of  molasses,  at  $28.75 
a  barrel ;  and  64  sacks  of  coffee,  at  $47.50  a  sack.  Place 
the  purchases  in  bill  form. 


SOLUTION. 


JOSEPH  ALLEN, 


1879. 


SPRINGFIELD,  ILL.,  Jan.  2,  1879. 
Bought  of  SETH  WARD. 


Jan. 

2 

To  30  bl.  flour,           @  $  3.60  a  bl. 

108 

00 

,, 

2 

,,    48  ,,    mess  pork,   ,,     16.25    ,, 

780 

00 

,, 

2 

,,    16  boxes  candles,  ,,       3.50    ,,  box 

56 

00 

„ 

2 

„    23  bl.  molasses,      „     28.75    ,,  bl. 

661 

25 

» 

2 

,,    64  sacks  coffee,       „     47.50    ,,  sack 

8040 

00 

84645 

25 

2.  At  St.  Louis,  March  1,  1879,  Chester  Snyder  bought 
of  Thomas  Glenn,  4  Ib.  of  tea,  at  40  ct. ;  21  Ib.  of  butter, 
at  21  ct.;  58  Ib.  of  bacon,  at  13  ct.;  16  Ib.  of  lard,  at  9 
ct.;  30  Ib.  of  cheese,  at  12  ct.;  4  Ib.  of  raisins,  at  20  ct.; 
and  9  doz.  of  eggs,  at  15  ct.  Place  these  purchases  in  the 
form  of  a  receipted  bill?  $20.74. 

66.  A  Statement  of  Account  is  a  written  form  renr 
dered  to  a  customer,  showing  his  debits  and  credits  as  they 
appear  on  the  books.  The  following  is  an  example : 


JOHN  SMITH, 


1880. 


CINCINNATI,  Feb.  2,  1880. 
In  Account  with  VAN  ANTWERP,  BRAGG  &  Co. 


Jan. 


10 


To  525  McGuffey's  Revised  First  Readers,  @  16c. 
,,     50  Ray's  New  Higher  Arithmetics,        ,,  75c. 

Cr. 
By  Cash 
i  ,f   Merchandise 

84 
37 

50 
75 

50 

:75 

~^r 

20 
12 

121 
32 

.  $88 

38  JRAY'S  HIGHER  ARITHMETIC. 

3.  James  Wilson  &  Co.  bought  of  the  Alleghany  Coal 
Co.,  March  2,  1880,  five  hundred  tons  of  coal,  at  $2.75  a 
ton,  and  sold  the  same  Company  during  the  month,  as 
follows:  March  3d,  14  barrels  of  flour,  at  $6.55  a  barrel; 
March  10th,  6123  pounds  of  sugar,  at  8c.  a  pound ;  they 
also  paid  them  on  account,  on  March  15th,  cash,  $687.50. 
Make  out  a  statement  of  account  in  behalf  of  the  Alleghany 
Coal  Co.  under  date  of  April  1,  1880.  $105.96. 


CONTRACTIONS  IN  MULTIPLICATION. 

CASE   I. 

67.     When  the  multiplier  is  a  composite  number. 

A  Composite  Number  is  the  product  of  two  or  more 
whole  numbers,  each  greater  than  1,  called  its  factors. 
Thus,  10  is  a  composite  number,  whose  factors  are  2  and 
5;  and  30  is  one  whose  factors  are  2,  3,  and  5. 

PROBLEM. — At  7  cents  a  piece,  what  will  6  melons  cost? 

ANALYSIS. — Three  times  2  times  OPERATION. 

are  6  times.     Hence,  it  is  the  same  7  cents,  cost  of  1  melon. 

to  take  2  times  7,  and  then   take  2 

this  product  3  times,  as  to  take  6  14  cents,  cost  of  2  melons. 

times  7.     The  same  may  be  shown  3 

of  any  other  composite  number.  ^  cents?  cost  of  6  melOns. 

Rule. — Separate  the  multiplier  into  two  or  more  factors. 
Multiply  first  by  one  of  the  factors,  then  this  product  by  another 
factor,  and  so  on  till  each  factor  has  been  used  as  a  multiplier. 
The  last  product  wiU  be  the  result  required. 

EXAMPLES  FOR  PRACTICE. 

1.  At  the  rate  of  37  miles  a  day,  how  far  will  a  man 
walk  in  28  days?  1036  miles. 


MUL  TIPLICA  T1ON.  39 

2.  Sound    moves    about  1130   feet  per  second:    how   far 
will  it  move  in  54  seconds?  61020  feet. 

3.  If   an    engine    travel    at    an    average    speed    of    25 
miles  an   hour,  how  far  can    it  travel  in  a  week,  or  168 
hours?  4200  miles. 

CASE  II. 

68.  When  the  multiplier  is  1  with  ciphers  annexed, 
as  10,  100,  1000,  etc. 

DEMONSTRATION.  —  By  the  principles  of  Notation  (Art.  43), 
placing  one  cipher  on  the  right  of  a  number,  changes  the  units 
into  tens,  the  tens  into  hundreds,  and  so  on,  and,  therefore,  multiplies 
the  number  by  10. 

Annexing  two  ciphers  to  a  number  changes  the  units  into  hun- 
dreds, the  tens  into  thousands,  and  so  on,  and,  therefore,  multiplies 
the  number  by  100.  Annexing  three  ciphers  multiplies  the  number 
by  1000,  etc. 

Rule.  —  Annex  to  the  multiplicand  as  many  ciphers  as  there 
are  in  the  multiplier  ;  the  result  will  be  the  required  product. 

EXAMPLES  FOR  PRACTICE. 

1.  Multiply  743  by  10.  Ans.  7430. 

2.  Multiply  375  by  100.  Ans.  37500. 

3.  Multiply  207  by  1000,  Ans.  207000. 

CASE    III. 

69.  When  ciphers  are  on  the  right  in  one  or  both 
factors. 

PROBLEM.—  Find  the  product  of  5400  by  130. 

OPERATION. 
5400 

SOLUTION.  —  Find  the  product  of  54  by  13,  1  30 

and    then    annex    three    ciphers  ;  that    is,    as  162 

many  as  there  are  on  the    right  in  both    the  54 


factors.  702000 


40  HAY'S  HIGHER  ARITHMETIC. 

ANALYSIS. — Since  13  times  54  —  702,  it  follows  that  13  times  54 
hundreds  (5400)  =  702  hundreds  (70200);  and  130  times  5400=10 
times  13  times  5400  ==  10  times  70200  ==  702000. 

Rule. — Multiply  as  if  there  were  no  ciphers  on  the  right  in 
the  numbers;  then  annex  to  the  product  as  many  ciphers  as 
there  are  on  the  right  in  both  the  factors. 

EXAMPLES  FOR  PRACTICE. 

1.  15460  X  3200.  Ans.  49472000. 

2.  30700  X  5904000.  Ans.  181252800000. 

CASE    IV. 

70.  When  the  multiplier  is  a  little  less  or  a  little 
greater  than  10,  100,  1000,  etc. 

PROBLEM.— Multiply  3046  by  997. 

ANALYSIS. — Since    997   is    equal    to    1000  OPERATION. 

diminished  by  3,  to  multiply  by  it  is  the  same  3046 

as  to  multiply  by  1000  (that  is,  to  annex  3  997 

ciphers)  and  by  3,  and  take  the  difference  of  3046000 

the  products ;  and  the  same  can  be  shown  in  9138 

any  similar  case.  3036862 

NOTE.— Where  the  number  is  a  little  greater  than  10,  100,  1000, 
etc.,  the  two  products  must  be  added. 

Rule. — Annex  to  the  multiplicand  as  many  ciphers  as  there 
are  figures  in  the  multiplier;  multiply  the  multiplicand  by  the 
difference  between  the  multiplier  and  100,  1000,  etc.,  and  add 
or  subtract  the  smaller  result  as  the  multiplier  is  greater  or  less 
than  100,  1000,  etc. 

EXAMPLES  FOR  PRACTICE. 

1.  7023  X  99.  Ans.  695277. 

2.  16642  X  996.  Ans.  16575432. 

3.  372051  X  1002.  Ans.  372795102. 


MULTIPLICATION.  41 


CASE    V. 

71.  When  one  part  taken  as  units,  in  the  multi- 
plier, is  a  factor  of  another  part  so  taken. 

PROBLEM. -Multiply  387295  by  216324. 

SOLUTION. — Commence   with    the    3    of  OPERATION. 

the  multiplier,  and  obtain  the  first  partial  387295 

product,  1161885 ;  then  multiply  this  prod-  216324 

uct  by  8,  which  gives  the  product  of  the  1161885 

multiplicand  by  24  at  once  (since  8  times  9295080 

3  times  any  number   make  24  times  it).         83655720 
Set  the  right-hand  figure  under  the  right-         83781203580 
hand  hgure  4  of   the  multiplier  in  use. 

Multiply  the  second  partial  product  by  9,  which  gives  the  product 
of  the  multiplicand  by  216  (since  9  times  24  times  a  number  make 
216  times  that  number).  Set  the  right-hand  figure  of  this  partial 
product  under  the  6  of  the  multiplicand ;  and,  finally,  add  to  obtain 
the  total  product. 

Rule. — 1.  Multiply  the  multiplicand  by  some  figure  or  figures 
of  the  multiplier,  which  are  a  factor  of  one  or  more  parts  of  the 
multiplier. 

2.  Multiply  this  partial  product   by  a  factor  of  some  other 
figure    or   figures   of  the  multiplier,  and  write   the   right-hand 
figure  thus  obtained  under  the  right-hand  figure  of  the  multiplier 
thus  used. 

3.  Continue  thus  until  the  entire  multiplier  is  used,  and  then 
add  the  partial  products. 

EXAMPLES  FOR  PRACTICE. 

1.  38057  X  48618.  Am.  1850255226. 

2.  267388  X  14982.  Ans.  4006007016. 

3.  481063  X  63721.  An*.  30653815423. 

4.  66917X849612.  Am.  56853486204. 

5.  102735  X  273162.  Ans.  28063298070. 

6.  536712  X  729981.  Am.  391789562472. 

H.  A.  4. 


42 


RAY'S  HIGHER  ARITHMETIC. 


1.  Definitions. 


2.  Terms.... 


3.  Sign. 

4.  Principles. 


5.  Operation • 


6.  Rule. 

7.  Proof. 

8.  Applications. 

9.  Contractions. 


Topical  Outline. 
MULTIPLICATION. 


»1.  Multiplicand. 
2.  Multiplier. 
3.  Partial  Product. 
4.  Product. 


1.  Writing  Numbers. 

2.  Drawing  Line  Beneath. 

3.  Finding  Partial  Products. 

4.  Drawing  a  Line  Beneath  Partial  Products. 

5.  Adding  the  Partial  Products. 


YL  DIVISION. 

72.  1.  Division    is    the   process  of   finding   how    many 
times  one  number  is  contained  in  another  ;  or, 

2.  Division  is  a  short  method   of  making   several    sub- 
tractions of  the  same  number. 

3.  Division  is  also  an  operation  in  which  are  given  the 
product  of  two  factors,  and  one  of  the  factors,  to  find  the 
other  factor. 

73.  The   product  is  the   Dividend;   the    given  factor  is 
the   Divisor;  and  the   required  factor  is  the  Quotient.     The 
Remainder   is    the    number    which    is    sometimes  left   after 
dividing. 

NOTE.  —  Dividend  signifies  to  be  divided.    Quotient  is  derived  from 
the  Latin  word  quoties,  which  signifies  how  often. 

PROBLEM.  —  How   many  times   is   24    cents   contained  in 
73  cents? 

SOLUTION.  —  Twenty  -four     cents  OPERATION. 

from  73  cents  leaves  49  cents;  24  73  cents. 

cents  from  49  cents  leaves  25  cents  ;  24 

24  cents  from  25  cents  leaves  1  cent.  49  cents  remaining. 

Here,  24  cents  is  taken  3  times  24 

from  (out  of)  73  cents,  and  1    cent  ^  cents  remaining. 

remains  ;  hence,  24  cents  is  contained  24 

m  73  cents  3  times,  with  a  remainder  -J  cent  remaining 
of  1  cent. 

74.  The  divisor  and  quotient  in  Division,  correspond  to 
the  factors  in  Multiplication,  and  the  dividend  corresponds  to 
the  product.     Thus  : 


Dividend  {  \  Divisors  and  Quotients. 

^    1  O  -f-  o  :=  O    ) 


(43) 


44  RAY'S  HIGHER  ARITHMETIC. 

75.  There    are    three    methods    of  expressing    division ; 
thus, 

12-1-3,  \2-,  or    3)12. 

Each  indicates  that  12  is  to  t  be  divided  by  3. 

PRINCIPLES. — 1.    When  the  dividend   and  divisor   are   like 
numbers,  the  quotient  is  abstract. 

2.  When  the  divisor  is  an  abstract  number,  the  quotient  is 
like  the  dividend. 

3.  The  remainder  is  like  the  dividend. 

4.  The   dividend  is  equal  to  the  product  of  the  quotient  by 
the  divisor,  plus  the  remainder. 

76.  Multiplication  is  a  short  method  of  making  several 
additions  of  the  same  number ;  Division  is  a  short  method 
of  making  several  subtractions  of  the  same  number ;  hence, 
Division  is  the  reverse  of  Multiplication. 

77.  All    problems    in    Division    are    divided    into    two 
classes : 

1.  To  find  the  number  of  equal  parts  of  a  nwnber. 

2.  To  divide  a  number  into  equal  parts. 

78.  Two  methods  are  employed  in   solving   problems  in 
Division:  Long  Division,  when  the  work  is  written  in  full 
in   solving   the  problem;    and    Short   Division,   when    the 
result  only  is  written,  the  work  being  performed  in  the  mind. 

The  following  illustrates  the  methods : 
PROBLEM.— Divide  820  by  5. 

LONG   DIVISION.  SHORT   DIVISION. 

5)820(164  Quotient.  5)820 

5  164  Quotient. 

32  tens. 

i 

Both  operations  are  performed  on  the  same 
20  units.  principle.     In  the  first,  the  subtraction  is  writ- 

2J)  teii ;  in  the  second,  it  is  performed  mentally. 


DIVISION.  45 


LONG  DIVISION. 

PROBLEM. — Divide  $4225  equally  among  13  men. 

SOLUTION. — As   13   is  not  contained  OPERATION. 
in  4   (thousands),  therefore,    the   quo- 
tient has  no  thousands.     Next,  take  42  g'g  ».•§    *g  £% 
(hundreds)   as  a  partial  dividend;  13  £5.23    ££3 
is  contained  in  it  3  (hundreds)  times ;  13)4225)325 
after  multiplying  and  subtracting,  there  3  9  hundreds, 
are  3  hundreds  left.     Then  bring  down 

2  tens,  and  32  tens  is  the  next  partial  26  tens, 

dividend.      In  this,  13  is  contained   2  65 

(tens)  times,  with  a  remainder  of  6  tens.  6  5  units. 
Lastly,  bringing  down  the  5  units,  13 
is   contained   in   65   (units)    exactly  5 
(units)  times.     The  entire  quotient  is  3  hundreds  2  tens  and  5  units. 

This  may  be  further  shown  by  separating  the  dividend  into  parts, 
each  exactly  divisible  by  13,  as  follows: 

DIVISOR.  DIVIDEND.  QUOTIENT. 

13)3900  +  260  +  65(300+20  +  5 
3900 

~+260 
+  260 

+  65 
+  65 

Rule  for  Long  Division. — 1.    Draw    curved  lines  on  the 
right  and  left  of  the  dividend,  placing  the  divisor  on  the  left. 

2.  Find  how  often  the  divisor  is  contained    in   the  left-hand 
figure,  or  figures,  of  the  dividend,  and  write  the  number  in  the 
quotient  at  the  right  of  the  dividend. 

3.  Multiply  the  divisor  by  this  quotient  figure,  and  write  the 
product  under  that  part    of   the    dividend  from    which  it  was 
obtained. 

4.  Subtract  this  product  from  the  figures  above  it ;  to  the  re- 
mainder bring  down  the  next  figure  of  the  dividend,  and  divide 
as   beforey   until  all    the  figures   of   the    dividend   are    brought 
down. 


46     -  RAY'S  HIGHER  ARITHMETIC. 

5.  If  at  any  time  after  a  figure  is  brought  down,  the  number 
thus  formed  is  too  small  to  contain  the  divisor,  a  cipher  must 
be  placed  in  the  quotient,  and  another  figure  brought  down,  after 
which  divide  as  before. 

6.  If  there  is  a  final  remainder  after  the  last  division,  place 
flie  divisor  under  it  and  annex  it  to  the  quotient. 

PROOF. — Multiply  the  Divisor  by  the  Quotient,  and  to 
this  product  add  the  Remainder,  if  any ;  the  sum  is  equal 
to  the  Dividend  when  the  work  is  correct. 

NOTES. — 1.  The  product  must  never  be  greater  than  the  partial 
dividend  from  which  it  is  to  be  subtracted;  if  so,  the  quotient 
figure  is  too  large,  and  must  be  diminished. 

2.  The  remainder  after   each  subtraction  must  be  less  than  the 
divisor ;  if  not,  the  last  quotient  figure  is  too  small,  and  must  be 
increased. 

3.  The  order  of  each  quotient  figure  is  the  same  as  the  lowest 
order  in  the  partial  dividend  from  which  it  was  derived* 

EXAMPLES  FOR  PRACTICE. 

1.  1004835  —  33.  Am.  30449f|. 

2.  5484888  —  67.  Ans.  81864. 

3.  4326422  —  961.  Ans.  4502. 

4.  1457924651  -1204.  Ans.  1210900|{Hft. 

5.  65358547823  -f  2789.  Am.  234344022GT%V 

6.  33333333333-^-5299.  Ans.  629Q495^\\. 

I.  245379633477  —  1263.  Ans.  194283161fif|-. 

8.  555555555555  —  123456.  Am.  45000281*$£$r. 

9.  555555555555  —  654321.  Am.  849056f££fff 

In  the  following,  multiply  A  by  itself,  also  B  by  itself: 
divide  the  difference  of  the  products  by  the  sum  of  A  and  B. 

A.  B. 

10.  2856  3765.  Am.  909. 

II.  33698  42856.  Ans.  9158. 
12.  47932  152604.  Am.  104672. 


DIVISION.  47 

In  the  following,  multiply  A  by  itself,  also  B  by  itself: 
divide  the  difference  of  the  products  by  the  difference  of 
A  and  B. 

A.  B. 

13.  4986  5369.  Ans.  10355. 

14.  3973  4308.  Ans.  8281. 

15.  23798  59635.  Ans.  83433. 

16.  47329  65931.  Ans.  113260. 

17.  If  25  acres  produce  1825  bushels  of  wheat,  how  much 
is  that  per  acre  ?  73  bushels. 

18.  How  many  times  1024  in  1048576?  1024  times. 

19.  How  many  sacks,  each  containing  55  pounds,  can  be 
filled  with  2035  pounds  of  flour?  37  sacks. 

20.  How   many  pages  in  a  book  of  7359  lines,  each  page 
containing  37  lines?  198ff  pages. 

21.  In  what  time  will  a  vat  of  10878  gallons  be  filled,  at 
the  rate  of  37  gallons  an  hour?  294  hours. 

22.  In  what  time  will  a  vat  of  3354  gallons  be  emptied, 
at  the  rate  of  43  gallons  an  hour?  78  hours. 

23.  The   product   of  two  numbers   is  212492745;  one  is 
1035;  what  is  the  other?  205307. 

24.  What  number  multiplied  by  109,  with  98  added  to 
the  product,  will  give  106700?  978. 


SHOET  DIVISION. 

PROBLEM. — How  often  is  2  cents  contained  in  652  cents? 

SOLUTION. — Two  in  6  (hundreds)  is  contained  3      OPERATION. 
(hundreds)    times ;    2    in   5  (tens)  is   contained  2         2 )  652 
(tens)  times,  with  a  remainder  of  1  (ten) ;  lastly,  1  326 

(ten)  prefixed  to  2  makes  12,  and  2  in  12  (units) 
is  contained  6  times,  making  the  entire  quotient  326. 

REMARKS. — Commence  at  the  left  to  divide,  so  that  if  there  is  a 
remainder  it  may  be  carried  to  the  next  lower  order. 


48  RAY'S  HIGHER  ARITHMETIC. 

By  the  operation  of  the  rule,  the  dividend  is  separated  into  parts 
corresponding  to  the  different  orders.  Having  found  the  number 
of  times  the  divisor  is  contained  in  each  of  these  parts,  the  sum  of 
these  must  give  the  number  of  times  the  divisor  is  contained  in  the 
whole  dividend.  Analyze  the  preceding  dividend  thus : 

652  =  600  +  40  +  12 
2  in  600  is  contained  300  times. 
2  in     40  is  contained     20  times. 
2  in      1  2  is  contained        6  times. 
Hence,  2  in  652  is  contained  326  times. 


Rule  for  Short  Division. — 1.  Write  the  divisor  on  the 
left  of  the  dividend  with  a  curved  line  between  them,  and  draw 
a  line  directly  beneath  the  dividend.  Begin  at  the  left,  divide 
successively  each  figure  or  figures  of  the  dividend  by  the  divisor, 
and  set  the  quotient  beneath. 

2.  Whenever  a  remainder  occurs,  prefix   it   to  the  figure  in 
the  next  lower  order,  and  divide  as  before. 

3.  If  the  figure,  except  the  first,  in  any  order  does  not  con- 
tain the  divisor,  place  a  cipher  beneath  it,  prefix  it  to  the  figure 
in  the  next  lower  order,  and  divide  as  before. 

4.  If  there  is  a  remainder    after    dividing   the    last  figure, 
place  the  divisor  under  it  and  annex  it  to  the  quotient. 

PROOF. — The  same  as  in  Long  Division. 


EXAMPLES  FOR  PRACTICE. 

1.  Divide  512653  by    5.  Ans.  102530f. 

2.  Divide  534959  by     7.  Ans.  76422f 

3.  Divide  986028  by     8.  Ans.  123253f. 

4.  Divide  986974  by  11.  Ans.  89724|f 

5.  At  $6  a   head,   how   many   sheep   can  be  bought  for 
$222?  37  sheep. 

6.  At   $5   a   barrel,  how   many   barrels   of   flour  can  be 
bought  for  $895?  179  barrels. 


DIVISION.  49 


CONTRACTIONS  IN  DIVISION. 

CASE    I. 

79.  When  the  divisor  is  a  composite  number. 

This  case  presents  no  difficulty  except  when  remainders 
occur. 

PROBLEM.  —  Divide  217  by  15. 

SOLUTION.  —  15  =  3  X  5,  hence  217  -i-  3  =  72  and  1  remainder  ; 
72  -r-  5  =14  and  2  remainder.  Dividing  217  by  3,  the  quotient  is 
72  threes,  and  1  unit  remainder.  Dividing  by  5,  the  quotient  is  14 
(fifteens),  and  a  remainder  of  2  threes;  hence  the  quotient  is  14, 
and  the  true  remainder  is  2  X  3  +  1  =  7. 

Rule.  —  1.  Divide  the  dividend  by  one  factor  of  the  divisor,  and 
divide  this  quotient  by  another  factor,  and  so  on,  till  each  factor 
has  been  used;  the  last  quotient  will  be  the  required  result. 

2.  Multiply  each  remainder  by  all  of  the  divisors  preceding 
the  one  which  produced  it.  The  sum  of  the  products,  plus  the 
first  remainder,  will  be  the  true  remainder. 

BEMARK.  —  This  rule  is  not  much  used. 

CASE    II. 

80.  When  the  divisor  is  1  with  ciphers  annexed. 

This  case  presents  no  difficulty.     Proceed  thus: 

PROBLEM.—  Divide  23543  by  100. 

OPERATION. 
SOLUTION.—  1  1  0  0  )235|43 


Rule.  —  Cut  off  as  many  figures  in  the  dividend  as  there  are 
ciphers  in  the  divisor;  the  figures  cut  off  will  be  the  remainder, 
and  the  other  figure  or  figures  the  quotient. 

H.  A.  5. 


50  RAY'S  HIGHER  ARITHMETIC. 

CASE   III. 

81.  When  ciphers  are  on  the  right  of  the  divisor. 
PROBLEM.— Divide  3846  by  400. 

SOLUTION.— To    divide  by  400  is  the  OPERATION. 

same  as  to  divide  by  100  and  then  by  4  4)00)38146 
(Art.  79).     Dividing  by  100  gives  38,  and  9   Quotient, 

46   remainder   (Art.  80);  then,  dividing  200  +  46  =  246,  Kern, 
by  4  gives  9,  and  2  remainder:  the  true 
remainder  is  2  X  100  +  46  =  246  (Art.  79). 

Rule. — 1.  Cut  off  the  ciphers  at  the  right  of  the  divisor, 
and  as  many  figures  from  the  right  of  the  dividend. 

2.  Divide  the  remaining  part  of  the  dividend  by  the  remain- 
ing part  of  the  divisor. 

3.  Annex   to   the   remainder   the  figures   cut   off,    and   thus 
obtain  the  true  remainder. 

ARITHMETICAL  SIGNS. 

82.  If  a  number  be  multiplied,  it  is  simply  repeated  as 
many  times  as  there  are  units  in  the  multiplier;  if  a  num- 
ber be  divided,  it  is  simply  decreased  by  the  divisor  as  many 
times  as  there  are  units  in  the  quotient.     It  is  thus  evident, 
that  Addition  and  Subtraction  are  the  fundamental  concep- 
tions in   all  the  operations  of  Arithmetic;    and,  hence,  all 
numbers  may  be  classified  as  follows : 

1.  Numbers  to  be  added;  or,  positive  numbers. 

2.  Numbers  to  be  subtracted;  or,  negative  numbers. 

83.  Positive  numbers  are  distinguished  by  the  sign  -J-, 
negative  numbers  by  the  sign  • — ;  thus,  +  8  is  a  positive  8, 
and  — •  8  a  negative  8. 

KEMARK. — When  a  number  is  preceded  by  no  sign,  as,  for 
example,  the  number  4  in  the  first  of  the  following  exercises,  it  is 
to  be  considered  positive. 


ARITHMETICAL  SIGNS.  51 

84.  The  signs  X  and  -f-  do  not  show  whether  their  results 
are  to  be  added  or  to  be  subtracted ;  they  simply  show  what 
operations  are  to  be  performed  on  the  positive  or  negative 
numbers  which  they  follow. 

Thus,  in  the  statement,  +  12  —  5X2,  the  sign  X  shows  that  5  is 
to  be  taken  twice,  but  it  does  not  show  what  is  to  be  done  with  the 
resulting  10;  that  is  shown  by  the  — .  We  are  to  take  two  5's  from 
12.  So,  in  18  +  9  -f-  3,  the  sign  -=-  shows  that  9  is  to  be  divided  by 
3 ;  what  is  to  be  done  with  the  quotient,  is  shown  by  the  +  before 
the  9. 

85.  In  every  such  numerical  statement,  the  +  or  the  — 
must  be   understood  to  affect  the  whole  result  of  the  opera- 
tions indicated  between  it  and  the  next  +  or  — ,  or  between  it 
and  the  close  of  the  expression. 

Thus,  in  5  +  7  X  2  X  9  —  2  X  6,  the  +  indicates  the  addition  of 
126,  not  of  7  only ;  and  the  —  indicates  the  subtraction  of  12.  The 
same  meaning  is  conveyed  by5+(7X2X9)  —  (2X6). 

86.  When  the  signs  X  and  -f-  occur  in  succession,  they 
are   to  have  their  particular  effects  in  the  exact  order  of 
their  occurrence. 

Thus,  we  would  indicate  by  96  -r- 12  X  4,  that  the  operator  is  first 
to  divide  by  12,  and  then  multiply  the  quotient  by  4.  The  result 
intended  is  32,  not  2;  if  the  latter  were  intended,  we  should  write 
96  -r-  (12  X  4).  Usage  has  been  divided  on  this  point,  however. 

EEMARK.— It  will  be  observed  that  in  no  case  can  the  sign  X 
or  ~  affect  any  number  before  the  preceding  +  or  — ,  or  beyond  the 
following  +  or  — . 

EXERCISES. 

1.  4X3  +  7X2  —  9X3+6X4  —  3X3  =  ? 

SOLUTION. +  4X3  =  12,  7X2  =  14,  —9X3=  -27,  6X4  =  24, 
—  3X3  =  —  9.  Grouping  and  adding  according  to  the  signs,  we 
have,  12  +  14  +  24  =  50 ;  and  —  27  —  9  ==  —  36.  Therefore,  50  —  36 
=  14,  Ans. 

2.  2X2  —  1X2  —  2X2  —  5X3—5X3  —  4X2  — 

4x2  — 8X3— 5X2  — 9X3  — 7X2  — 12  X4  —  7X 


52  RAY'S  HIGHER  ARITHMETIC. 

Soi>UTioN.+  4  —2  —  4  —  1 5  —  15  —  8  —  8  —  24  — 10  —  27  — 14  — 
1.g  — 14.  Grouping  and  adding,  we  have,  4  — 189  —  — 185,  Ans. 

3.  21  -f-  3  X  7  —  1X1-7-1X4-7-2  + 18 -=-3x6-7- 
(2X2) +  (4  —  2  +  6  —  7)  X  4x6-^8  =  ?  59. 

EEMARK. — Whenever  several  numbers  are  included  within  the 
marks  of  parenthesis,  brackets,  or  vinculum,  they-  are  regarded  as 
(me  number.  Note  the  advantage  of  this  in  example  3. 

4.  .16x4-1-8  —  7  +  48-^16—3—7x4x0x9X16  + 
24x6-f-48  —  4  x  9-M2  =  ?  1. 

5.  (16-M6X96-T-8  —  7  —  5  +  3)  X[ (27-J-9 )  -7-8- 
1]  + (91 -J-13X7— 45  —  3)X9=?  9. 


GENEEAL  PEINCIPLES. 

87.     The  following  are  the  General  Principles  of  Mul- 
tiplication and  Division. 

PRINCIPLE  I. — Multiplying  either  factor  of  a  product,  multi- 
plies the  product  by  the  same  number. 

Thus,  5X4  =  20,  and  5X4X2  =  40,  whence  20X2  =  40. 

II. — Dividing  either  factor  of  a  product,  divides  the  product 
by  the  same  number. 

Thus,  5X4  =  20,  and  5X4-5-2  =  10,  whence  20-5-2  =  10. 

III. — Multiplying  one  factor  of  a  product,  and  dividing  the 
other  factor  by  the  same  number,  does  not  alter  the  product. 

Thus,  6X4  =  24,  and  6X2X4-5-2  =  24,  whence  6X2X2  = 
24. 

IV.—  Multiplying  the   dividend,   or  dividing  the   divisor,  by 
any  number,  multiplies  the  quotient  by  that  number. 

If  24  be  the  dividend  and  6  the  divisor,  then  4  is  the  quotient; 
hence  24X2n-6  =  8,  and  24 -s- (6 -5- 2)  =  8. 


CONTRACTIONS.  53 

V. — Dividing   the  dividend,  or  multiplying   the   divisor,  by 
any  number,  divides  the  quotient  by  that  number. 

Thus,  if  24  be  the  dividend  and  6  the  divisor,  then  24  -^  2  =  12, 

and  12  -*-  6  =  2 ;  whence  24  -r-  (6  X  2)  =  2.     Therefore,  4  +  2  =  2. 

VI. — Multiplying  or  dividing  both  dividend  and   divisor  by 
the  same  number,  does  not  change  the  quotient. 

Thus,  24  X  2  =  48,   and  6  X  2  =  12 ;    consequently,  48  -*- 12  =  4, 
and  24-5-6  =  4? 


CONTRACTIONS   IN    MULTIPLICATION    AND   DIVISION. 

CASE    I. 

88.     To  multiply  by  any  simple  part  of  100, 1000,  etc, 

NOTE. — Let  the  pupil  study  carefully  the    following  table  of 
equivalent   parts : 

PARTS  OF  100.  PARTS  OF  1000. 

121  =  |  of  100.  125    =  •§•  of  1000. 

16|  =  |  of  100.  166f  =  £  of  1000. 

25    =  J  of  100.  250    =  £  of  1000. 

=  t  of  100.  3331  =  £  of  1000. 

=  f  of  100.  375  =  f  of  1000. 

621  =  I  of  100.  625  —I  of  1000. 

66f  =  |  of  100.  666|  —  |  of  1000. 

75  =f  of  100.  750  =|  of  1000. 

871  =  £  of  100.  875  =  $  of  1000. 

PROBLEM. — Multiply  246  by  87^.       OPERATION. 

24600 

SOLUTION.  —  Since    87 J   is   £   of    100,  7 

annex  two  ciphers  to   the    multiplicand,  g  \i72200 

which   multiplies    it  by   100,    and    then  . _ 

take  J  of  the  result. 

Rule. — Multiply  by  100,    1000,  etc.,   and  take  such  a  part 
of  the  result  as  the  multiplier  is  of  100,  1000,  etc. 


54  RA  Y>  S  HIGHER  ARITHMETIC. 

EXAMPLES  FOR  PRACTICE. 

1.  *22  X  33i  An*.  14066|. 

2.  6564  X  62J.  ^w*.  410250. 

3.  10724  X  16|.  Ans.  178733|. 

CASE  II. 

89.  To   multiply  by  any  number  whose  digits  are 
all  alike. 

PROBLEM.— Multiply  592643  by  66666. 

SOLUTION.— Multiply  592643  by  99999  (Art.  70),  the  product  is 
59263707357 ;  take  f  of  this  product,  since  6  is  f  of  9 ;  the  result 
is  39509138238. 

Rule. — Multiply  as  if  the  digits  were  9'«,  and  take  such  a 
part  of  the  product  as  the  digit  is  of  9. 

EXAMPLES  FOR  PRACTICE. 

1.  451402  X  3333.  Ans.  1504522866. 

2.  281257  X  555555.  Ans.  156253732635. 

3.  630224  X  4444000.  Ans.  2800715456000. 

/ 

CASE    III. 

90.  To   divide   by  a  number  ending  in  any  simple 
part  of  100,  1000,  etc. 

PROBLEM.— Divide  6903141128  by  21875. 

SOLUTION. — Multiply  both  by  8  and  4  successively.  The  divisor 
becomes  700000,  and  the  dividend  220900516096,  while  the  quotient 
remains  the  same.  (Art.  87,  VI.)  Performing  the  division  as  in 
Art.  81,  the  quotient  is  315572,  and  remainder  116096.  The  remain- 
der being  a  part  of  the  dividend,  has  been  made  too  large  by  the 
multiplication  by  8  and  4,  and  is,  therefore",'  reduced  to  its  true 
dimensions  by  dividing  by  8  and  4.  This  gives  3628  for  the  true 
remainder. 


CONTRACTIONS.  55 

Rule. — Multiply  both  dividend  and  divisor  by  such  a  number 
as  will  convert  the  final  figures  of  the  divisor  into  ciphers,  and 
then  divide  the  former  product  by  the  latter. 

NOTES.— -1.  If  there  be  a  remainder,  it  should  be  divided  by  the 
multiplier,  to  get  the  true  remainder. 

2.  The  multiplier  is  3,  4,  6,  etc.,  according  as  the  final  portion 
of  the  divisor  is  thirds,  fourths,  sixths,  etc.,  of  100,  1000. 


EXAMPLES  FOR  PRACTICE. 

1.  300521761^225.  Ans. 

2.  1510337264^43750.  Ans. 

3.  22500712361-^1406250.  Ans. 

4.  620712480 -=-20833$-.  Quot.  29794.     Bern.  4146|. 

5.  742851692 -r-29161-  Quot.  254692.     Bern. 


GENERAL   PROBLEMS. 

NOTE. — Let  the  pupil  make  a  special  problem  under  each  general 
problem,  and  solve  it. 

1.  When  the  separate  cost   of  several  things    is  given, 
how  is  the  entire  cost  found? 

2.  When  the  sum  of  two  numbers,  and  one  of  them,  are 
given,  how  is  the  other  found? 

3.  When    the  less  of   two    numbers  and    the    difference 
between  them  are  given,  how  is  the  greater  found  ? 

4.  When  the  greater  of  two  numbers  and  the  difference 
between  them  are  given,  how  is  the  less  found? 

5.  When  the   cost   of  one  article  is  given,  how  do  you 
find  the  cost  of  any  number  at  the  same  price? 

6.  If  the  total   cost   of  a  given   number  of  articles  of 
equal  value  is  stated,  how   do  you   find   the  value  of  one 
article  ? 

7.  When  the  divisor  and  quotient  are  given,  how  do  you 
find  the  dividend? 


56  RA  Y'S  HIGHER  ARITHMETIC. 

8.  How  do  you  divide  a  number  into  parts,  each  contain- 
ing a  certain  number  of  units? 

9.  How  do  you  divide  a  number  into  a  given  number  of 
equal  parts? 

10.  If  the  product  of  two  numbers,  and  one  of  them,  are 
given,  how  do  you  find  the  other? 

11.  If  the  dividend  and  quotient  are  given,  how  do  you 
find  the  divisor? 

12.  If  you  have  the  product  of  three  numbers,  and  two 
of  them  are  given,  how  do  you  find  the  third? 

13.  If  the  divisor,   quotient,  and  remainder   are  given, 
Low  do  you  find  the  dividend  ? 

14.  If  the   dividend,  quotient,  and  remainder  are  given, 
how  do  you  find  the  divisor? 


MISCELLANEOUS  EXERCISES. 

1.  A  grocer  gave  153  barrels  of  flour,  worth  $6  a  bar- 
rel, for  54  barrels  of  sugar :  what  did  the  sugar  cost  per 
barrel?  $17. 

2.  When  the  divisor  is  35,  quotient  217,  and  remainder 
25,  what  is  the  dividend?  7620. 

3.  What  number  besides  41  will  divide  4879  without  a 
remainder?  119. 

4.  Of  what   number   is    103    both   the  divisor   and  the 
quotient?  10609. 

5.  What   is  the  nearest  number  to  53815,  that  can  be 
divided  by  375  without  a  remainder?  54000. 

6.  A  farmer  bought  25  acres  of  land  for  $2675 :  what  did 
19  acres  of  it  cost?  $2033. 

7.  I  bought  15  horses,  at  $75  a  head :  at  how  much  per 
head  must  I  sell  them  to  gain  $210  ?  $89. 

8.  A  locomotive  has  391  miles  to  run  in  11  hours:  after 
running  139  miles  in  4  hours,  at  what  rate  per  hour  must 
the  remaining  distance  be  run?  36  miles. 


MISCELLANEO  US  EXER  CISES.  57 

9.  A  merchant  bought   235  yards  of   cloth,  at   $5  per 
yard :  after  reserving  12  yards,  what  will  he  gain  by  selling 
the  remainder  at  $7  per  yard  ?  $386. 

10.  A  grocer  bought   135  barrels  of  pork  for  $2295 ;  he 
sold  83  barrels  at  the  same  rate  at  which  he  purchased,  and 
the  remainder  at  an  advance  of  $2  per  barrel :  how  much 
did  he  gain?  $104. 

11.  A  drover  bought  5  horses,  at  $75  each,  and    12  at 
$68   each;    he  sold   them  all   at  $73   each:   what   did    he 
gain  ?  $50. 

At  what  price  per  head  must  he  have  sold  them  to  have 
gained  $118?  $77. 

12.  A  merchant  bought  3  pieces  of  cloth  of  equal  length, 
at  $4   a  yard;  he  gained  $24  on  the  whole,  by  selling  2 
pieces    for   $240:     how    many    yards    were    there    in   each 
piece?  18  yards. 

13.  If  18  men  can  do  a  piece  of  work   in   15  days,  in 
how  many  days  will  one  man  do  it?  270  days. 

14.  If  13  men  can  build  a  wall  in  15  days,  in  how  many 
days  can  it  be  done  if  8  men  leave?  39  days. 

15.  If  14  men  can  perform  a  job  of  work  in  24  days,  in 
how  many  days  can  they  perform  it  with  the  assistance  of 
7  more  men?  16  days. 

16.  A  company  of  45  men  have  provisions  for  30  days : 
how  many  men  must  depart,    that  the  provisions  .may  last 
the  remainder  50  days?  18  men. 

17.  A   horse  worth   $85,  and  3  cows  at   $18  each,  were 
exchanged  for  14  sheep  and  $41  in  money :    at  how  much 
each  were  the  sheep  valued?  $7. 

18.  A  drover  bought  an  equal  number  of  sheep  and  hogs 
for  $1482:    he  gave  $7    for    a   sheep,  and    $6    for  a  hog: 
what  number  of  each  did  he  buy?  114. 

19.  A  trader  bought  a  lot  of  horses  and  oxen  for  $1260; 
the  horses  cost  $50,  and  the  oxen  $17,  a  head;  there  were 
twice  as  many  oxen  as  horses :    how  many  were  there   of 
each?  15  horses  and  30  oxen. 


58 


RAY'S  HIGHER  ARITHMETIC. 


20.  In    a   lot  of   silver    change,  worth    1050    cents,   one 
seventh  of  the  value  is  in  25-cent  pieces ;  the  rest  is  made 
up  of  10-cent,  5-cent,  and  3-cent  pieces,  of  each  an  equal 
number :  how  many  of  each  coin  are  there  ? 

Of  25-cent  pieces,  6 ;  of  the  others,  50  each. 

21.  A  speculator  had  140  acres  of  land,  which  he  might 
have   sold  at  $210    an  acre,  and  gained  $6300;    but  after 
holding,  he   sold  at  a  loss   of  $5600 :    how  much    an  acre 
did  the  land  cost  him,  and  how  much  an  acre  did  he  sell 
it  for?  $165,  cost;  and  $125,  sold  for. 


Topical  Outline. 
DIVISION. 

1.  Dividend. 

2.  Divisor. 

3.  Quotient. 

4.  Remainder. 

'  1.  Writing  the  Numbers. 

2.  Drawing  Curved  Lines. 

3.  Finding  Quotient  Figure. 

4.  Multiplying  Divisor  and  Writing  Product. 

5.  Drawing  Line. 

6.  Subtracting. 

7.  Annexing  Lower  Order. 

8.  Repeating  the  Process  from  3. 
te9.  Writing  Remainder. 

6  Rules..  |  LonS  ^vision. 

7  Proof. 

8.  Applications. 


1.  Definitions. 

2.  Terms 

3.  Sign. 

4.  Principles. 


5.  Operation 


I  Short  Division. 


9.  Contractions /  Division. 

\  Multiplication  and  Division. 

f  1..0f  Numbers... /Positive' 

10.  Arithmetical  Signs.  \  *  NeSative- 

I  2.  Of  Operation... /Multiplying. 
I  Dividing. 

11.  General  Principles. 

12.  Applications. 


VII.  PEOPERTIES  OF  NUMBERS. 


DEFINITIONS. 

91.  1.  The  Properties  of  Numbers  are  those  qualities 
which  belong  to  them. 

2.  Numbers    are    classified    (1),  as   Integral,   Fractional, 
and  Mixed  (Art.  22)  ;   (2),  as  Abstract  and  Concrete  (Art. 
21);  (3),  Prime  and  Composite;   (4),  Even  and  Odd;   (5), 
Perfect  and  Imperfect. 

3.  An  integer  is  a  whole  number;  as,  1,  2,  3,  etc. 

4.  Integers  are   divided  into  two  classes— prime  numbers 
and  composite  numbers. 

5.  A  prime  number  is  one  that  can  be  exactly  divided 
by  no   other  whole  number  but   itself  and  unity,  (1) ;  as, 
1,  2,  3,  5,  7,  11,  etc. 

6.  A   composite   number  is   one   that  can  be   exactly 
divided    by  some    other    whole  number   besides    itself  and 
unity;  as,  4,  6,  8,  9,  10,  etc. 

REMARK. — Every  composite  number  is  the  product  of  two  or 
more  other  numbers,  called  its  factors  (Art.  60). 

7.  Two   numbers  are  prime  to  each  other,  when  unity  is 
the  only  number  that  will  exactly  divide  both;  as,  4  and  5. 

REMARK. — Two  prime  numbers  are  always  prime  to  each  other: 
sometimes,  also,  two  composite  numbers;  as,  4  and  9. 

8  An  even  number  is  one  which  can  be  divided  by  2 
without  a  remainder ;  as,  2,  4,  6,  8,  etc. 

9.  An  odd  number  is  one  which  can  not  be  divided  by 
2  without  a  remainder;  as,  1,  3,  5,  7,  etc. 

REMARK. — All  even  numbers  except  2  are  composite  ;  the  odd 

numbers  are  partly  prime  and  partly  composite. 

(59) 


60  RA  Y'S  HIGHER  ARITHMETIC. 

10.  A  perfect  number  is  one  which  is  equal  to  the  sum 
of  all  its   divisors ;    as,  6  — 1  +  2  +  3;    28  =  1  +  2  +  4  + 
7  +  14. 

11.  An  imperfect  number  is  one  not  equal  to  the  sum 
of  all  its  divisors.     Imperfect  numbers  are  Abundant  or  De- 
fective:    Abundant  when  the  number  is  less  than  the  sum 
of  the  divisors ;    as,  18,  less  than  1  +  2  +  3  +  6  +  9;  and 
Defective  when  the  number  is  greater  than  the  sum ;    as, 
16,  greater  than  1+2  +  4  +  8. 

12.  A  divisor  of  a  number,  is  a  number  that  will  exactly 
divide  it. 

13.  One  number  is  divisible  by  another  when  it  contains 
that  other  without  a  remainder ;  8  is  divisible  by  2. 

14.  A  multiple  of  a  number  is  the  product  obtained  by 
taking   it  a  certain  number  of  times;    15  is  a  multiple  of 
5,  being  equal  to  5  taken  3  times;  hence, 

1st.  A  multiple  of  a   number   can   always   be   divided  by  it 
'without  a  remainder. 

2d.  Every  multiple  is  a  composite  number. 

15.  Since  every  composite   number    is   the    product    of 
factors,  each  factor  must  divide  it  exactly;    hence,  every 
lactor  of  a  number  is  a  divisor  of  it. 

16.  A  prime  factor  of  a  number  is  a  prime  number  that 
will  exactly  divide  it :  5  is  a  prime  factor  of  20 ;  while  4  is 
a  factor  of  20,  not  a  prime  factor ;  hence, 

1st.   The  prime  factors  of  a  number  are  all  the  prime  num- 
bers that  urill  exactly  divide  it. 

EXAMPLE. — 1,  2,  3,  and  5  are  the  prime  factors  of  30. 

2d.  Every  composite  number  is  equal  to   the  product   of  all 
its  prime  factors. 

EXAMPLE. — All  the  prime  factors  of  15  are  1,  3,  and  5 ;  and  1  X 
3  X  5  =  15. 

17.  Any  factor  of  a  number  is  called   an   aliquot  part 
of  it. 

EXAMPLE. — 1,  2,  3,  4,  and  6,  are  aliquot  parts  of  12. 


FACTORING.  61 


FACTOEING. 

92.  Factoring  is  resolving  composite  numbers  into  fac- 
tors ;  it  depends  on  the  following  principles  and  propositions. 

PRINCIPLE  1. — A  factor  of  a  number  is  a  factor  of  any 
multiple  of  that  number. 

DEMONSTRATION. — Since  6  =  2  X  3,  therefore,  any  multiple  of 
6  =  2  X  3  X  some  number ;  hence,  every  factor  of  6  is  also  a  factor 
of  the  multiple.  The  same  may  be  proved  of  the  multiple  of  any 
composite  number. 

PRINCIPLE  2. — A  factor  of  any  two  numbers  is  also  a  factor 
of  their  sum. 

DEMONSTRATION. — Since  each  of  the  numbers  contains  the  factor 
a  certain  number  of  times,  their  sum  must  contain  it  as  often  as 
both  the  numbers ;  2,  which  is  a  factor  of  6  and  10,  must  be  a  factor 
of  their  sum,  for  6  is  3  twos,  and  10  is  5  twos,  and  their  sum  is  3 
twos  +  5  twos  =  8  twos. 

93.  From  these  principles  are  derived  the  six  following 
propositions : 

PROP.  I. — Every  number  ending  with  0,  2,  4,  6,  or  8,  is 
divisible  by  2. 

DEMONSTRATION. — Every  number  ending  with  a  0,  is  either  10  or 
some  number  of  tens;  and  since  10  is  divisible  by  2,  therefore,  by 
Principle  1st,  Art.  92,  any  number  of  tens  is  divisible  by  2. 

Again,  any  number  ending  with  2,  4,  6,  or  8,  may  be  considered 
as  a  certain  number  of  tens  plus  the  figure  in  the  units'  place ;  and 
since  each  of  the  two  parts  of  the  number  is  divisible  by  2,  there- 
fore, by  Principle  2d,  Art.  92,  the  number  itself  is  divisible  by  2; 
thus,  36  =  30  +  6  =  3  tens -f  6;  each  part  is  divisible  by  2;  hence, 
36  is  divisible  by  2. 

Conversely,  No  number  is  divisible  by  2,  unless  it  ends  with 
0,  2,  4,  6,  or  8. 

PROP.  II. — A  number  is  divisible  by  4,  when  the  number 
denoted  by  its  two  right-hand  digits  is  divisible  by  4. 


62  JRA  Y>  S  HIGHER  ARITHMETIC. 

DEMONSTRATION.— Since  100  is  divisible  by  4,  any  number  of 
hundreds  will  be  divisible  by  4  (Art.  92,  Principle  1st) ;  and  any 
number  consisting  of  more  than  two  places  may  be  regarded  as  a 
certain  number  of  hundreds  plus  the  number  expressed  by  the 
digits  in  tens7  and  units'  places  (thus,  384  is  equal  to  3  hundreds 
-f-  84) ;  then,  if  the  latter  part  (84)  is  divisible  by  4,  both  parts, 
or  the  number  itself,  will  be  divisible  by  4  (Art.  92,  Prin.  2d). 

Conversely,  No  number  is  divisible  by  4,  unless  the  number 
denoted  by  its  two  right-hand  digits  is  divisible  by  4. 

PROP.  III. — A  number  ending  in  0  or  5  is  divisible  by  5. 

DEMONSTRATION. — Ten  is  divisible  by  5,  and  every  number  of  two 
or  more  figures  is  a  certain  number  of  tens,  plus  the  right-hand 
digit ;  if  this  is  5,  both  parts  of  the  number  are  divisible  by  5,  and, 
hence,  the  number  itself  is  divisible  by  5  (Art.  92,  Prin.  2d). 

Conversely,  No  number  is  divisible  by  5,  unless  it  ends  in 
0  or  5. 

PROP.  IV. — Every  number  ending  in  0,  00,  etc.,  is  divis- 
ible by  10,  100,  etc. 

DEMONSTRATION. — If  the  number  ends  in  0,  it  is  either  10  or  a 
multiple  of  10 ;  if  it  ends  in  00,  it  is  either  100,  or  a  multiple  of  100, 
and  so  on ;  hence,  by  Prin.  1st,  Art.  92,  the  proposition  is  true. 

PROP.  V. — A  composite  number  is  divisible  by  the  product 
of  any  two  or  more  of  its  prime  factors. 

DEMONSTRATION.— Since  2  X  3  X  5  =  30,  it  follows  that  2  X  3 
taken  5  times,  makes  30;  hence,  30  contains  2X3(6)  exactly  5 
times.  In  like  manner,  30  contains  3  X  5  (15)  exactly  2  times,  and 
2X5  (10),  exactly  3  fimes. 

Hence,  If  any  even  number  is  divisible  by  3,  it  is  also 
divisible  by  6. 

DEMONSTRATION. — An  even  number  is  divisible  by  2 ;  and  if  also 
by  3,  it  must  be  divisible  by  their  product  2X3,  or  6. 

PROP.  VI. — Every  prime  number,  except  2  and  5,  ends 
with  1,  3,  7,  or  9. 

DEMONSTRATION. — This  is  in  consequence  of  Props.  I.  and  III. 


FACTORING.  63 

PROP.  VII. — Any  integer  is  divisible  by  9  or  by  3,  if  the 
sum  of  its  digits  be  thus  divisible. 

94.  To  find  the  prime  factors  of  a  composite 
number. 

PROBLEM. — Find  the  prime  factors  of  42. 

SOLUTION. — 42  is  divisible   by  2,  and  21  is  OPERATION. 
divisible  by  3   or  7,  which  is  found  by  trial ;  2)42 

hence,  the  prime  factors  of  42  are  2,  3,  7,  /.  2  X  3)21 

3X7-42.  — 

Rule. — Divide  the  given  number  by  any  prime  number  that 
will  exactly  divide  it;  divide  the  quotient  in  like  manner,  and 
so  continue  until  the  quotient  is  a  prime  number;  the  several 
divisors  and  the  last  quotient  are  the  prime  factors. 

KEMARKS. — 1.  Divide  first  by  the  smallest  prime  factor. 

2.  The  least  divisor  of  any  number  is  a  prime  number  ;  for,  if  it 
were   a  composite   number,  its  factors,  which  are  less  than  itself, 
would  also  be  divisors   (Art.  92),   and    then   it  would  not  be   the 
least  divisor.     Therefore,  the  prime  factors  of  any  number  may  be 
found   by  dividing  it   first  by  the   least  number  that  will  exactly 
divide  it,  then  dividing  this  quotient  in  like  manner,  and  so  on. 

3.  Since  1  is  the  factor  of  every  number,  either  prime  or  com- 
posite, it  is  not  usually  specified  as  a  factor. 


Find  the  prime  factors  of: 

L  45.  Ans.  3,  3,  5. 

2.  54.         Ans.  2,  3,  3,  3. 

3.  72.     Am.  2,  2,  2,  3,  3. 


4.  75.  Ans.  3,  5,  5. 

5.  96.      Ans.  2,  2,  2,  2,  2,  3. 

6.  98.  Ans.  2,  7,  7. 


7.  Factor  210.  Ans.  2,  3,  5,  7. 

8.  Factor  1155.  Ans.   3,  5,  7,  11. 

9.  Factor  10010.  Ans.  2,  5,  7,  11,  13. 

10.  Factor  36414.  Ans.  2,  3,  3,  7,  17,  17. 

11.  Factor  58425.  Ans.  3,  5,  5,  19,  41. 


64  RAY'S  HIGHER  ARITHMETIC. 

95.  The  prime  factors  common  to  several  numbers  may 
be    found    by  resolving    each   into   its  prime   factors,  then 
taking  the  prime  factors  alike  in  all. 

Find  the  prime  factors  common  to: 

1.  42  and  98.  Am.  2,  7. 

2.  45  and  105.  Am.  3,  5. 

3.  90  and  210.  Ans.  2,  3,  5. 

4.  210  and  315.  Ans.  3,  5,  7. 

96.  To  find  all  the  divisors  of  any  composite  num- 
ber. 

Any  composite  number  is  divisible,  not  only  by  each  of 
its  prime  factors,  but  also  by  the  product  of  any  two  or 
more  of  them  (Art.  93,  Prop.  V.)  ;  thus, 

42  =  2X3X7;  and  all  its  divisors  are  2,  3,  7,  and 
2  X  3,  2  X  7,  and  3  X  7;  or,  2,  3,  7,  6,  14,  21.  Hence, 

Rule. — Resolve  the  number  into  its  prime  factors,  and  then 
form  from  these  factors  all  the  different  products  of  which  they 
will  admit;  the  prime  factors  and  their  products  will  be  all  the 
divisors  of  the  given  number. 

Find  all  the  divisors: 

1.  Of  70.  Ans.  2,  5,  7,  and  10,  14,  35. 

2.  Of  196.  Ans.  2,  7,  and  4,  14,  28,  49,  98. 

3.  Of  231.  Ans.  3,  7,  11,  and  21,  33,  77. 

4.  Of  496 ;  and  name  the  properties  of  496. 

GKEATEST    COMMON    DIVISOR 

97.  A  common  divisor  (C.  D.)  of  two  or  more  num- 
bers, is  a  number  that  exactly  divides  each  of  them. 

98.  The  greatest  common  divisor  (G.  C.  D.)  of  two 
or  more  numbers  is  the  greatest  number  that  exactly  divides 
each  of  them. 


GREATEST  COMMON  DIVISOR.  65 

PRINCIPLES. — 1.  Every  prime  factor  of  a  number  is  a 
divisor  of  that  number. 

2.  Every  product  of  two  or  more  prime  factors  of  a  number, 
is  a  divisor  of  that  number. 

3.  Every  number  is  equal  to  the  continued  product  of  all  its 
prime  factors. 

4.  A  divisor  of  a   number  is   a    divisor  of  any  number  of 
times  that  number. 

5.  A  common  divisor  of  two  or  more  numbers  is  a  divisor 
of  their  sum,  and  also  of  their  difference. 

6.  The  product  of  all  the  prime  factors,  common  to  two  or 
more  numbers,  is  their  greatest  common  divisor. 

7.  The  greatest  common  divisor  of  two  numbers,  is  a  divisor 
of  their  difference. 

To  FIND  THE  GREATEST  COMMON  DIVISOR. 

CASE    I. 

99.  By  simple  factoring. 

PROBLEM.— Find  the  G.  C.  D.  of  30  and  105. 

OPERATION. 

30  =  2  X  3  X  5.    |    3  x  5  =  15,  G.  C.  D. 
105  =  3X5X7.    ) 

DEMONSTRATION. — The  product  3  X  5  is  a  divisor  of  both  the 
numbers,  since  each  contains  it,  and  it  is  their  greatest  common 
divisor,  since  it  contains  all  the  factors  common  to  both. 

PROBLEM.— Find  the  G.  C.  D.  of  36,  63,  144,  and  324. 

OPERATION. 


SOLUTION.  — 

3 
3 

36,   63,   1 

44,    324 

12,    21, 

48,    108 

4,       7, 

16,      36 

/.  3X3  =  ' 

},  G.  C.  D. 

Rule. — Resolve  the  given  numbers  into  their  prime  factors, 
and  take  the  product  of  the  factors  common  to  all  the  numbers. 

H.  A.  6. 


66  RAY* 8  HIGHER  ARITHMETIC. 

Find  the  greatest  common  divisor : 

1.  Of  30  and  42.  Ans.  2x3  =  6. 

2.  Of  42  and  70.  Ans.  2x7==  14. 

3.  Of  63  and  105.  Ans.  3  X  7  =  21. 

4.  Of  66  and  165.  Ans.,  3  X  11—33. 

5.  Of  90  and  150.  Am.  2  X  3  X  5  =  30. 

6.  Of  60  and  84.  Ans.  2  X  2  X  3  =  12. 

7.  Of  90  and  225.  Ans.  3x3x5  =  45. 

8.  Of  112  and  140.  Ans.  2  X  2  X  7  =  28. 

9.  Of  30,    45,  and  75.  Ans.  3  X  5  =  15. 

10.  Of  84,  126,  and  210.  Ans.  2x3x7  =42. 

11.  Of  16,  40,  88,  and  96.  Ans.  2x2x2  =  8. 

12.  Of  21,  42,  63,  and  126.  Ans.  3  X  7  =  21. 

CASE    II. 

100.  By  successive  divisions. 

PROBLEM.— Find  the  G.  C.  D.  of  348  and  1024. 

OPERATION. 

DEMONSTRATION.— If  348         348)  1  0  2  4  (J2_ 
will    divide    1024,  it  is  the  696 

G.  C.  D. ;    but   it   will    not  328  )  3  4  8  ( 1 

divide  it.  ~  3  2  8  ~ 

If  328  (Art  98,  Prin.  5,)  — ,  3  2  g  fl  6 

will   divide    348,  it    is    the  —         ' 

G.  C.  D.  of  328,  348,  and 
1024 ;  but  it  will  not  divide 
348  and  1024  exactly.  *  2  ° 


If  20  will  divide  328  (by  8)20(2 

same  process  of  reasoning),  1  6 

it  is  the  G.  C.  D.;  but  there  G.  C.  D.   4)8 

is  a  remainder  of  8 ;  hence,  2 

if  8  will  divide  20,  it  is  the 
G.  C.  D. ;  but  there  is  also  a  remainder  of  4. 

Now,  4  divides  8  without  a  remainder.  Therefore,  4  is  the 
greatest  number  that  will  divide  4,  8,  20,  328,  348?  and  1024,  and  is 
the  G.  C.  D. 


GREATEST  COMMON  DIVISOR.  67 

Rule. — Divide    the     greater    number    by   the    less,    and  the 

divisor  by  the  remainder,  and   so  on;   always   dividing  the  last 

divisor   by  the    last   remainder,  till   nothing  remains;    the  last 
divisor  will  be  the  greatest  common  divisor  sought. 

NOTE. — A  condensed  form  of  operation    CONDENSED  OPERATION. 
may  be  used  after  the   pupils  are  familiar  348 

with  the  preceding  process. 


328 


REMARK. — To  find  the  greatest  common 
divisor  of  more  than  two  numbers,  find  the 
G.  C.  D.  of  any  two ;  then  of  that  G.  C.  D.,  16 

and  any  one  of    the  remaining    numbers,  4 

and  so  on  for  all  of  the  numbers ;  the  last 
C.  D.  will  be  the  G.  C.  D.  of  all  the  numbers. 


1024 
696 


328 
320 


8 


1 
16 

2 
2 


REMARK. — If  in  any  case  it  be  obvious  that  one  of  the  numbers  has 
a  prime  factor  not  found  in  the  other,  that  factor  may  be  suppressed 
by  division  before  applying  the  rule.  Thus,  let  the  two  numbers  be 
715  and  11011.  It  is  plain  that  the  prime  5  divides  the  first  but  not 
the  second  ;  and  since  .that  prime  can  be  no  factor  of  any  common 
divisor  of  the  two,  their  G.  C.  D.  is  the  same  as  the  G.  C.  D.  of  143 
and  11011. 

Find  the  greatest  common  divisor  of: 

1.  85  and  120.  Ans.  5. 

2.  91  and  133.  Ans.  1. 

3.  357  and  525.  Ans.  21. 

4.  425  and  493.  Ans.  17. 

5.  324  and  1161.  Ans.  27. 

6.  589  and  899.  Ans.  31. 

7.  597  and  897.  Am.  3. 

8.  825  and  1287.  Ans.  33. 

9.  423  and  2313.  Ans.  9. 

10.  18607  and  24587.  Ans.  23. 

11.  105,  231,  and  1001.  Ans.  7. 

12.  165,  231,  and  385.  Ans.  11. 

13.  816,  1360,  2040,  and  4080.  Ans.  136. 

14.  1274,  2002,  2366,  7007,  and  13013.  Ans.  91. 


68  RAY'S  HIGHER  ARITHMETIC. 


LEAST  COMMON  MULTIPLE. 

101.  A   common    multiple    (C.   M.)    of   two  or  more 
numbers,  is  a  number  that  can  be  divided  by  each  of  them 
without  a  remainder. 

102.  The  least   common  multiple  (L.  C.  M.)  of  two 
or  more  numbers,  is  the  least   number  that  is  divisible  by 
each  of  them  without  a  remainder. 

PRINCIPLES. — 1.    A   multiple  of  a  number  is  divisible  by 
that  number. 

2.  A  multiple  of  a  number  must  contain  all  of  the  prime 
factors  of  that  number. 

3.  A  common  multiple  of  two   or  more  numbers  is  divisible 
by  each  of  those  numbers. 

4.  A  common  multiple  of  two  or  more  numbers  contains  all 
of  the  prime  factors  of  each  of  those  numbers. 

5.  The  least  common  multiple  of  two  or  more  numbers  must 
contain  all  of  the  prime  factors  of  each  of  those  numbers^  and 
no  other  factors. 

6.  If  two  or  more  numbers  are  prime  to  each   other,  tJieir 
continued  product  is  their  least  common  multiple. 

To  FIND  THE  LEAST  COMMON  MULTIPLE. 

CASE  I. 

103.  By  factoring  the  numbers  separately. 

PROBLEM.— Find  the  L.  C.  M.  of  10,  12,  and  15. 

SOLUTION. — Resolve  each  OPERATION. 

number  into  its  prime  f ac-  10=2X5 

tors.     A  multiple  of  10  con-  12=2X2X3 

tains  the  prime  factors  2  and  15  =  3X5 

5 ;  of  12,  the  prime  factors  .'.  L.  C.  M.  =  2  X  2  X  3  X  5  =  6  0. 
2,  2,  and  3  ;  of  15,  the  prime 
factors  3  and  5.     But  the  L.  C.  M.  of  10,  12,  and  15  must  contain 


LEAST  COMMON  MULTIPLE.  69 

all  of  the  different  prime  factors  of  these  numbers,  and  no  other 
factors  ;  hence,  the  L.  C.  M.  =  2  X  2  X3  X  5  =  60  (Art.  102,  Prin. 
2  and  5). 

Rule. — Resolve  each  number  into  its  prime  factors,  and 
then  take  the  continued  product  of  all  the  different  prime  factors, 
using  each  factor  the  greatest  number  of  times  it  occurs  in  any 
one  of  the  given  numbers. 

REMARKS. — 1.  Each  factor  must  be  taken  in  the  least  common  mul- 
tiple the  greatest  number  of  times  it  occurs  in  either  of  the  numbers. 
In  the  preceding  solution,  2  must  be  taken  twice,  because  it  occurs 
twice  in  12,  the  number  containing  it  most. 

2.  To  avoid  mistakes,  after  resolving  the  numbers  into  their 
prime  factors,  strike  out  the  needless  factors. 


Find  the  L.  C.  M.  of: 

1.  8,  10,  15.         Am.  120. 

2.  6,  9,  12.  Ans.  36. 

3.  12,  18,  24.         Ans.  72. 


4.  8,  14,  21,  28.       Ans.  168. 

5.  10,  15,  20,  30.       Ans.  60. 

6.  15,  30,  70,  105.    Ans.  210. 


CASE  II. 

104.    By  dividing  the  numbers  successively  by  their 
common  primes. 

PROBLEM.— Find  the  L.  C.  M.  of  10,  20,  25,   and  30. 
SOLUTION.  —  Write    the  OPERATION. 


20    25     30 


10    25     15 


numbers  as  in  the  margin.  2 

Strike  out  10,  because  it  is  ~ 

contained   in  20    and    30. 

Next,  divide  20  and  30  by  25 

the  prime  factor  2;  write  2X5X2X5X3=300  L.  C.  M. 
the  quotients  10  and  15, 

and  the  undivided  number  25  in  a  line  beneath.  Divide  these  num- 
bers by  the  common  prime  factor  5.  The  three  quotients — 2,  5,  3, 
are  prime  to  one  another ;  whence,  the  L.  C.  M.  is  the  product  of  the 
divisors  2,  5,  and  the  quotients  2,  5,  3.  By  division,  all  needless 
factors  are  suppressed. 


70  RAY'S  HIGHER  ARITHMETIC. 

Rule. — 1.  Write  the  numbers  in  a  horizontal  line;  strike 
out  any  number  that  will  exactly  divide  any  of  the  others;  divide 
by  any  prime  number  that  will  divide  two  or  more  of  them 
without  a  remainder;  write  the  quotients  and  undivided  num- 
bers in  a  line  beneath. 

2.  Proceed  with  this  line  as  before,  and  continue  the  opera- 
tion   till  no   number  greater   than  1  will   exactly  divide  two  or 
more  of  the  numbers. 

3.  Multiply  together  the  divisors  and  the  numbers  in  the  last 
line;    their  product  will  be  the  least  common  multiple  required. 

REMARK. — Prime  factors  not  obvious  may  be  found  by  Art.  100. 

Find  the  least  common  multiple  of: 

1.  6,  9,  20.  Ans.  180. 

2.  15,  20,   30.  Ans.  60. 

3.  7,  11,  13,  5.  Ans.  5005. 

4.  35,  45,  63,  70.  Ans.  630. 

5.  8,  15,  20,  25,  30.  Ans.  600. 

6.  30,  45,  48,  80,  120,  135.  Ans.  2160. 

7.  174,  485,  4611,  14065,  15423.  Ans.  4472670, 

8.  498,  85988,  235803,  490546.  Am.  244291908. 

9.  2183,  2479,  3953.  Ans.  146261. 
10.  1271,  2573,  3403.  Ans.  105493. 


SOME  PROPEKTIES  OF  THE  NUMBER  NINE. 

105.  Addition,  Subtraction,  Multiplication,  and  Division 
may  be  proved  by  "  casting  out  the  9's."  To  cast  the  9's 
out  of  any  number,  is  to  divide  the  sum  of  the  digits  by  9, 
and  find  the  excess. 

PROBLEM. — Find  the  excess  of  9's  in  768945, 

EXPLANATION. — Begin  at  the  left,  thus  :  7  +  6  are  13  ;  drop  the 
9;  4  +  8  are  12;  drop  the  9 ;  3  +  4  + 5  are  12  ;  drop  the  9;  the 
excess  is  3.  The  9  in  the  number  was  not  counted. 


CASTING   OUT  NINE> 

PRINCIPLE. — Any  number  divided  by  9, 
remainder  as  the  sum  of  its  digits  divided  by  9. 

ILLUSTRATION. 

700000  =  7  X  100000  =  7  X  ( 99999  +  1 )  =  7  X  99999  +  7 
60000  =  6  X  10000  =  6  X  (9999 +  1)  =6  X  9999  +  6 
8000  =  8X  1000  =  8X  (999  +  l)  =  8X  999  +  8 
900  =  9  X   100  =  9 X   (99  +  l)  =  9X   99  +  9 
40  =  4X    10  =  4X    (9  +  l)=4X    9  +  4 
I     5  =  5X     1=  5 

Whence,  7  X  99999  +  6  X  9999  +  8X999+  9  X  99  +  4X  9  +  7  + 
>  =  768945. 


768945=: 


SOLUTION.  —  An  examination  of  the  above  shows  that  768945  has 
been  separated  into  multiples  of  9,  and  the  sum  of  the  digits  com- 
posing the  number  ;  the  same  may  be  shown  of  any  other  number. 
There  can  be  no  remainder  in  the  multiples,  except  in  the  sum  of 
the  digits. 

PROOF  OF  ADDITION.  —  The  sum  of  the  excess  of  ffs  in  the 
several  numbers  must  equal  the  excess  of  9's  in  their  sum. 


ILLUSTRATION.  —  The  excesses  in  the  num- 
bers  are  8,  2,  4,  and  3,  and  the  excess  in  the 
sum  of  these  excesses  is  8.  The  excess  in  the 
sum  of  the  numbers  is  8,  the  two  excesses 
being  the  same,  as  they  ought  to  be  when  the 
work  is  correct. 


OPERATION. 
7352  8 

5834  2 

6241  4 

7302  JJ_ 

8 


26729 


PROOF  OF  SUBTRACTION.  —  The  excess  of  9's  in  the  minuend 
must  equal  the  sum  of  the  excess  of  ffs  in  the  subtrahend  and 

remainder. 

OPEB  ATION. 

ILLUSTRATION.—  As  the   min-        Minuend,       7640  8 

uend  is  the  sum  of  the   subtra-        Subtrahend,  1234  1 

Kemainder,   6406  7 


hend  and  remainder,  the  reason 
of  this  proof  is  seen  from  that  of 
Addition. 


PROOF  OF  MULTIPLICATION. — Find  the  excess  of  9's  in  the 
factors  and  in  the  product.  The  excess  of  ffs  in  the  product 
of  the  excesses  of  the  factors,  should  equal  the  excess  in  the  product 
of  the  factors  themselves. 


72  RAY'S  HIGHER  ARITHMETIC. 

ILLUSTRATION. — Multiply  835  by  76  ;  OPERATION. 

the  product  is  63460.     The  excess  in  the  835X76  =  63460 

multiplicand   is  7,  in    the  multiplier  4,  835,  excess  =7 

and  in   the  product  1 ;  the  two   former  76,       "       =4 

multiplied,  give  28 ;  and  the  excess  in  28  7X4=28,       "       =1 

is  also  1,  as  it  should  be.  63460,       "       =1 

PROOF  OF  DIVISION. — Find  the  excess  of  9's  in  each  of  the 
terms.  To  the  excess  of  9's  in  the  product  of  the  excesses  in  the 
divisor  and  quotient,  add  the  excess  in  the  remainder;  the  excess 
in  the  sum  should  equal  the  excess  in  the  dividend. 

OPERATION. 

ILLUSTRATION. — Divide  8915  by  25;  the  quo-        356,  excess  5 
tient  is  356,  and  the  remainder,  15.     The  excess  25,      "       7 

of  9's  in  the  divisor  is  7 ;   in   the  quotient,  5 ;  35 

their  product  is  35,  the  excess  of  which  is  8.  u 

The  excess  in  the  remainder  is  6.     6  +  8  =  14,  '       lt 

of  which  the  excess  is  5.     The  excess  of  9's  in 
the  dividend  is  also  5. 

14,      "       5 
8915,      "       5 


CANCELLATION. 

106.  Cancellation  is  the  process  of  crossing  out  equal 
factors  from  dividend  and  divisor. 

The  sign  of  Cancellation  is  an  oblique  line  drawn 
across  a  figure ;  thus,  $,  0,  $. 

PRINCIPLES. — 1.  Canceling  a  factor  in  any  number,  divides 
the  number  by  that  factor. 

2.  Canceling  a  factor  in  both  dividend  and  divisor,  does  not 
change  the  quotient.  (Art.  129,  III.) 

PROBLEM. — Multiply  75,  153,  and  28  together,  and  divide 
by  the  product  of  63  and  36. 


CANCELLATION.  73 

SOLUTION.  —  Indicate  the  operations  OPERATION. 

as   in  the   margin.     Cancel  4  out  of  2517  V 

28   and   36,    leaving   7    above    and   9 
below.     Cancel  this  7  out  of  the  divi- 


dend and  out  of  the  63  in  the  divisor,  PP  X  #0 

leaving  9  below.     Cancel   a  9  out  of  ^       ^ 

the   divisor   and   out  of   153  in    the  3 

dividend,  leaving  17   above.     Cancel  25X17 

3  out  of  9  and  75,  leaving  25  above  —  g  --  =  1  4  1  f 

and  3  below.    No  further  canceling  is 

possible;    the  factors    remaining   in  the  dividend  are   25  and  17, 

whose  product,  425,  divided  by  the  3  in  the  divisor,  gives  141|. 

Rule  for  Cancellation.  —  1.  Indicate  the  multiplications 
which  produce  the  dividend,  and  those,  if  any,  which  produce  the 
divisor. 

2.  Cancel  equal  factors  from  dividend  and  divisor  ;  multiply 
together  the  factors  remaining  in  the  dividend,  and  divide  the 
product  by  the  product  of  the  factors  left  in  the  divisor. 

NOTE.  —  If  no  factor  remains  in  the  divisor,  the  product  of  the 
factors  remaining  in  the  dividend  will  be  the  quotient  ;  if  only  one 
factor  is  left  in  the  dividend,  it  will  be  the  answer. 


EXAMPLES  FOR  PRACTICE. 

1.  How    many  cows,   worth    $24    each,    can  I  get  for  9 
horsey,  worth  $80  each?  30  cows. 

2.  I  exchanged  8  barrels  of  molasses,  each  containing  33 
gallons,    at   40  cents  a  gallon,  for    10    chests    of  tea,    each 
containing   24   pounds :    how    much   a    pound    did  the   tea 
cost  me?  44  cents. 

3.  How  many  bales  of  cotton,  of  400  pounds  each,  at  12 
cents  a  pound,  are  equivalent  to  6  hogsheads  of  sugar,  900 
pounds  each,  at  8  cents  a  pound?  9  bales. 

4.  Divide    15  X  24  X  112  X  40  X  10  by  25  X  36  X  56 
X  90.  34. 

H.  A.  7. 


74 


RA  Y'S  HIGHER   ARITHMETIC. 


Topical  Outline. 


PROPERTIES  OF  NUMBERS. 

Properties. 


1.  Definitions  ......      Numbers  Classified  ....... 


Integer  ......  - 

praction 

Mixed. 


Divisor;  Divisible  ;  -Multiple  ;  Factors; 
Prime  Factors;  Aliquot  Parts. 


Prime. 

Composite. 

Even. 

Odd. 

Perfect. 

Imperfect. 


T  1.  Definition. 

2.  Principles. 

2.  Factoring I    3-  Propositions. 

j   4.  Operations. 

5.  Rules, 

i    6.  Applications. 


3.  G.  C.  D.. 


1.  Definitions.... 

2.  Principles. 

3.  Operations..... 


1.  Common  Divisor. 

2.  Greatest  Common  Divisor. 

Case  I JRule- 

X  Applications. 

Case  II...  /  Rule. 

I  Applications. 


f  1.  Definitions /  L  Common  Multiple. 

\  2.  ~ 


4.  L.  C.  M i    2.  Principles. 

j    3.  Operations 

j    Case  II.. .\ 

.  Applications. 

,   1.  Addition. 

Some  Properties  of  the  No.  9..  {  l*  *     1C1P16'  2.  Subtraction. 

I  2.  Application  to...  4   „ 


Least  Common  Multiple. 

Case  I /  Rule' 

\  Applications. 

Case  II...  f 


6.  Can  cell:  L  t  Lou... . 


1.  Definition. 
12.  Sign,   /. 

3.  Principles. 

4.  Operation. 

5.  Rule. 

6.  Applications. 


3.  Multiplication 

4.  Division. 


VIII.   COMMON  FKACTIOm 

PEFINITIONS. 

107.  A  Fraction   is  an  expression  for  one  or  more  of 
the  equal  parts  of  a  divided  whole. 

108.  Fractions   are  divided   into  two  classes;   viz.,  com- 
mon  fractions  and  decimal  fractions. 

109.  A  Common  Fraction    is   expressed  by  two  num- 
bers, one   above  and  one  below  a  horizontal  line;  thus,  -f, 
which  is  read  two  thirds. 

110.  The  Denominator  is  the  number  below  the  line. 
It    shows  the   number   of    parts    into   which  the  whole  is 
divided,  and  thus  the  size  of  the  parts. 

111.  The  Numerator  is  the  number  above  the  line.     It 
shows  how  many  of  the  parts  are  taken. 

NOTE. — The  denominator  denominates,  or  names,  the  parts ;  the 
numerator  numbers  the  parts. 

112.  The  Terms  of  a  fraction  are  the    numerator  and 
the  denominator. 

ILLUSTRATION.— The  expression  f,  four  fifths,  shows  that  the  whole 
is  divided  into  five  equal  parts,  and  that  four  of  those  parts  are 
taken.  5  is  the  denominator,  4  is  the  numerator,  and  the  terms  of 
the  fraction  are  4  and  5. 

113.  Every  fraction  implies:    1.  That    a  number  is  di- 
vided ;  2.  That   the   parts   are  equal ;  3.  That  one  or  more 
of  the  parts  are  taken. 

114.  There  are  two  ways  of  considering  a  fraction  whose 
numerator  is  greater  than   1.     Four  fifths  may  be  4  fifths 
of  one  thing,  or  1  fifth  of  four  things ;  therefore, 

(75) 


76  RAY'S  HIGHER  ARITHMETIC. 

The  numerator  of  a  fraction  may  be  regarded  as  showing 
the  number  of  units  to  be  divided;  the  denominator,  the 
number  of  parts  into  which  the  numerator  is  to  be  divided ; 
the  fraction  itself  being  the  value  of  one  of  those  parts. 

Hence,  a  fraction  may  be  considered  as  an  indicated 
division  (Art.  75)  in  which, 

1.  The  dividend  is  the  numerator. 

2.  The  divisor  is  the  denominator. 

3.  The  quotient  is  the  fraction  itself. 

115.  The  value  of  a  fraction  is  its  relation  to  a  unit. 

116.  Fractions  are  divided  into  classes   with  respect  to 
their  value  and  form. 

(1).  As  to  value,  into  Proper,  Improper,  and  Mixed. 
(2).  As  to  form,  into  Simple,  Complex,  and  Compound. 

117.  A  Proper  Fraction  is  one  whose  numerator  is  less 
than  its  denominator ;  as,  ^ 

118.  An  Improper  Fraction  is  one  whose  numerator  is 
equal  to,  or  greater  than,  its  denominator;  as,  f. 

119.  A  Mixed  Number  is  a   number  composed   of  an 
integer  and  a  fraction ;  as,  3-| . 

120.  A    Simple    Fraction    is  a   single   fraction    whose 
terms  are  integral ;  as,  f ,  f ,  J-. 

121.  A   Complex    Fraction   is  one   which    has  one  or 

i  i 

both  of  its  terms  fractional ;  as,  -|r,  •§,  or  ff. 

2  3 

122.  A  Compound  Fraction  is  a  fraction  of  a  fraction ; 
as,  \  of  f . 

123.  An    Integer    may  be    expressed    as  a   fraction  by 
writing  1  under  it  as  a  denominator ;  thus,  •£,  which  is  read 
seven  ones. 


COMMON  FRACTIONS.  77 

124.  The  Reciprocal  of  a  number  is  1  divided  by  that 
number ;   thus,  the  reciprocal  of  5  is  |. 

125.  Similar  Fractions  are   those   that   have    the  same 
denominator ;  as,  f  and  f . 

126.  Dissimilar  Fractions  are  those  that  have  unlike 
denominators;  as,  f  and  f. 

REMARK. — The  word  "fraction"  is  from  the  Latin,  fmngo,  I 
break,  and  literally  means  a  broken  number.  In  mathematics, 
however,  the  word  "  fraction,"  as  a  general  term,  means  simply  the 
indicated  quotient  of  a  required  division. 


NUMERATION  AND  NOTATION  OF  FRACTIONS. 

127.  Numeration  of  Fractions   is   the  art  of  reading 
fractional  numbers. 

128.  Notation  of  Fractions   is  the  art  of  writing  frac- 
tional numbers. 

Rule  for  Reading  Common  Fractions. — Read  the  num- 
ber of  parts  taken  as  expressed  by  the  numerator,  and  then  the 
size  of  the  parts  as  expressed  by  tJie  denominator. 

EXAMPLE. — |-  is  read  seven  ninths. 

REMARK. — Seven  ninths  (J),  signifies  7  ninths  of  one,  or  J  of  7, 
or  7  divided  by  9. 

Rule  for  Writing  Common  Fractions. —  Write  the  num- 
ber of  parts ;  place  a  horizontal  line  below  it,  wider  which  write 
the  number  which  indicates  the  size  of  the  parts. 

Fractions  to  be  written  in  figures: 

Seven  eighths.  Four  elevenths.  Five  thirteenths.  One 
seventeenth.  Three  twenty-ninths.  Eight  twenty-firsts. 
Nine  forty-seconds.  Nineteen  ninety-thirds.  Thirteen  one- 
hundredths.  Twenty-four  one-hundred-and-fifteenths. 


78  RA  Y>  S  HIGHER  ARITHMETIC. 

129.     Since  a  fraction  is  an  indicated  division  (Art.  114)  ; 
therefore, 

PRINCIPLES. — I.  A  Fraction  is  multiplied, 
1st.  By  multiplying  the  numerator. 
2d.  By  dividing  the  denominator. 

II.  A  Fraction  is  divided, 

1st.  By  dividing  ilie  numerator. 
2d.  By  multiplying  the  denominator. 

III.  The  value  of  a  Fraction  is  not  changed, 

1st.  By  multiplying  both  terms  by  the  same  number. 
2d.  By  dividing  both  terms  by  the  same  number. 

KEMARK. — The  proof  of  I  is  found  in  Art.  87,  Principle  IV ;  the 
proof  of  II  is  in  Principle  V ;  and  of  III,  in  Principle  VI. 


KEDUCTION  OF  FKACTIONS. 

130.  Reduction  of  Fractions  consists  in  changing  their 
form  without  altering  their  value. 

CASE    I. 

131.  To  reduce  a  fraction  to  its  lowest  terms. 

REMARKS. — 1.  Reducing  a  fraction  to  lower  terms,  is  changing  it 
to  an  equivalent  fraction  whose  terms  are  smaller  numbers. 

2.  A  fraction  is  in  its  lowest  terms  when  the  numerator  and  de- 
nominator are  prime  to  each  other;  as,  f,  but  not  f. 

PROBLEM. — Reduc3  -f-g-  to  its  lowest  terms. 

SOLUTION. — Dividing  both  terms   by  the  FIRST  OPERATION. 

common  factor  2,  the  result  is  ^f ;  dividing  2  )  | §  =  }§ 

this  by  5  (129,  in\  the  result  is  f,  which  5 )  -if  —  f,  Ans. 
can  not  be  reduced  lower. 

Or,  dividing  at  once  by  10,  the  greatest  SECOND  OPERATION. 

common  divisor  of  both  terms,  the  result  is  10  )  f-J  =  J,  Ans. 
|,  as  before. 


REDUCTION  OF  FRACTIONS.  79 

Rule. — Reject  all  factors  common  to  botfi,  terms  of  the  fraction. 
Or,  divide  both  terms  of  the  fraction  by  their  greatest  common 
ditnsor. 


Reduce  to  their  lowest  terms : 


2.  |f.  Ans. 

3.  T\V  Ans. 

4.  .  Ans. 


Express  the  following  in  their  simplest  forms: 


5..       -  An*, 


6.  |ff  Am.  ff 

8.  |||.  Ans.  if. 

9. 
10.  -f 


11.  923  -*- 1491.   Ans. 

12.  890  —  1691. 


13.  2261-^-4123.     ^Irw. 

14.  6160  -f-  40480. 


CASE    II. 

132.     To  reduce  a  fraction  to  higher  terms. 

REMARK. — Reducing  a  fraction  to  higher  terms,  is  changing  it 
to  an  equivalent  fraction  whose  terms  are  larger  numbers. 

PROBLEM. — Reduce  f  to  fortieths. 

SOLUTION, — Divide  40  by  8,  the  quo-  OPERATION. 

tient  is  5  ;    multiply  both  terms   of   the  40-^-8  =  5 

given  fraction,  f,  by  5  (129,  ra),  and  the  5__5X5_  25    ^us 

result  is  ff ,  the  equivalent   fraction  re-  8  X  5 
quired. 

Rule. — Divide  the  required  denominator  by  the  denominator 
of  the  given  fraction ;  multiply  both  terms  of  the  given  fraction 
by  this  quotient ;  the  result  is  the  equivalent  fraction  required. 

1.  Reduce  ^  and  ^  to  ninety-ninths.  Ans.  |~J,  |-f. 

2.  Reduce  f,  -f-,  and  -^  to  sixty-thirds.     Ans.  ||,  ff,  ^. 

3.  Reduce  T8T,  ^  and  |^  to  equivalent  fractions  having 
6783  for  a  denominator.  Ans.  f|||,  |f||,  ffff 0 


80  RA  Y'S  HIGHER  ARITHMETIC. 


CASE    III. 

133.  To  reduce  a  whole  or  mixed  number  to  an 
improper  fraction. 

PROBLEM.  —  Eeduce  3f  to  an  improper  fraction;  to  fourths. 

SOLUTION.  —  In  1  (unit),  there  are  4  fourths;  in  3  (units),  there 
are  3  times  4  fourths,  =  12  fourths  :  and  12  fourths  +  3  fourths  = 
15  fourths. 

Rule.  —  Multiply  together  the  whole  number  and  the  denom- 
inator of  the  fraction:  to  the  product  add  the  numerator,  and 
write  the  sum  over  ilie  denominator. 

1.  In  $7f,  how  many  eighths  of  a  dollar?  Ans.  -%9-. 

2.  In  19f  gallons,  how  many  fourths?  Am.  ^  . 

3.  In  13f-J  hours,  how  many  sixtieths?  Ans.  -8^y-. 


Reduce  to  improper  fractions: 


4.  llf.  Ans. 

5.  15T8T.  Ans. 

6.  127-H-.          Ans. 


7.  109TV  Ans. 

Ans. 


9.  13ft.  Am.  iflp. 

REMARK. — To  reduce  a  whole  number  to  a  fraction  having  a 
given  denominator,  is  a  special  case  under  the  preceding. 

PROBLEM. — Eeduce  8  to  a  fraction  whose  denominator  is  7. 
SOLUTION. — Since  ^  equals  one,  8  equals  8  times  |,  or  -57?. 

CASE    IV. 

134.  To  reduce  an  improper  fraction  to  a  whole 
or  mixed  number. 

PROBLEM. — Reduce  ±/-  of  a  dollar  to  dollars. 

SOLUTION. — Since  5  fifths  make  1  dollar,  there  will  he  as  many 
dollars  in  13  fifths  as  5  fifths  are  contained  times  in  13  fifths ;  that 
is,  2|  dollars. 


REDUCTION  OF  FRACTIONS. 


81 


Rule. — Divide  the  numerator  by  the  denominator;  the  quotient 
will  be  the  whole  or  mixed  number. 

REMARK. — If  there  be  a  fraction  in  the  answer,  reduce  it  to  its 
lowest  terms. 


1.  In  -3-g7-  of  a  dollar,  how  many  dollars? 

2.  In  i|^  of  a  bushel,  how  many  bushels? 

3.  In  Jg%5-  of  an  hour,  how  many  hours? 

Reduce  to  whole  or  mixed  numbers: 


5. 

6. 


341  bu. 
hours. 


Ans.   1. 

7.  ifp. 

Ans.  105T7T. 

Am.  35. 

8.  -ff  a. 

Ans.  827ft- 

Ins.  88f. 

9.    .l^LG, 

Am.  509^. 

CASE   V. 

135.     To  reduce  compound  to  simple  fractions. 

PROBLEM. — Reduce  f  of  -f-  to  a  simple  fraction. 
SOLUTION. — £  of  |  =  -2V  >  1  °f  y 

~28'  *° 


OPERATION. 


28 


4X7      28 


Rule. — Multiply  the  numerators  together  for  the  numerator, 
and  ike  denominators  together  for  the  denominator  of  the  frac- 
tion, canceling  common  factors  if  they  occur  in  both  terms. 

REMARK. — Whole  or  mixed  numbers  must  be  reduced  to  im- 
proper fractions  before  applying  the  rule. 

PROBLEM. — Reduce  •§•  of  -ft  of  ^  to  a  simple  fraction. 

OPERATION. 

SOLUTION. — Indicate  the  work, 

and     employ     cancellation,     as          %  X    $    X    7 
shown     in     the     accompanying          3X  10X  HQ^^  ^'   ^nSt 
operation.  ^  . 


82  RAY'S  HIGHER  ARITHMETIC. 

Eeduce  to  simple  fractions: 

1.  i  of  f    of  f  Am.  f 

2.  I  of  |.    of  2f .  Ans.  f . 

3.  |  of  ||  of  2|.  .Am.  2. 

4.  i  of  |    of  3f.  Ans.  1£. 

5.  |  of  |-    of  ^  of  8f.  Ans.  3£. 

6.  |  of  |   of  -f-  of  |  of  4|.  -4ns.  |. 

7.  T8T  of  f   of  A  of  II  of  7i- 

8.  |f  of  A  of  A  of  ^  of  1¥V 

COMMON  DENOMINATOR 

136.  A  common  denominator  of  two  or  more  fractions, 
is  a  denominator  by  which  they  express  like  parts  of  a  unit. 

137.  The  least  common  denominator  (L.    C.    D.)  of 
two  or  more  fractions,  is   the    least   denominator   by  which 
they  can  express  like  parts  of  a  unit. 

PRINCIPLES — 1.  Only  a  common  multiple  of  different  de- 
nominators can  become  a  common  denominator. 

2.  Only  a  least  common  multiple  can  become  a  least  common 
denominator. 

CASE    VI. 

138.  To    reduce    fractions    to    equivalent   fractions 
having  a  common  denominator. 

PROBLEM. — Reduce  -^,  f ,  and  f  to  a  common  denominator. 

OPERATION. 

SOLUTION.— Since  2  X  3  X  4  =  24,  24  is  a  1X3X4^12 
common  multiple  of  all  the  denominators. 

The  terms  of  J  must  be  multiplied  by  3  X  4 ;  2  X2X4  __  1_6 
the  terms  of  f,  by  2  X  4 ;  and  the  terms  of  f ,  by  3X2X4  24 
2X3.  The  values  of  the  fractions  are  un-  3X^X3  =  18 
altered.  (129,  in.)  4X2X3  24 


DEDUCTION  OF  FRACTIONS.  83 

Rule.  —  Multiply  both  terms  of  each  fraction  by  the  denomina- 
tors of  the  oilier  fractions. 

REMARK.  —  Since  the  denominator  of  each  new  fraction  is  the 
product  of  the  same  numbers  —  viz.,  all  the  denominators  of  the 
given  fractions  —  it  is  unnecessary  to  find  this  product  more  than 
once.  The  operation  <  is  generally  performed  as  in  the  following 
example  : 

PROBLEM.  —  Reduce  |,  f,  and  ^  to  a  common  denominator. 

OPERATION. 

2X5X7  =  70,  common  denominator. 

1X5X7  =  35,  first  numerator.  i  =  $H 

3X2X7  =  42,  second  numerator.  f  =  f  f  (  Am. 

6  X  2  X  5  =  .6  0,  third  numerator.  f  =  f  £  J 

NOTE.  —  Mixed  numbers  and  compound  fractions  must  first  be 
reduced  to  simple  fractions  ;  the  lowest  terms  are  preferable. 

Reduce  to  a  common  denominator: 

i-  t,  I,  I-  A**-  ii  li  U- 

2-  \,    i    f  ^.   A«o>    T22\>   T\V 

3-  I,  f,  I-  -4«».  HI,  A2*,  IM- 

4-  i,  I,  I,  !•  Am.  fti,  iff,  ||o;  i|o. 

5.  f,  i  of  3},  |  of  f.  ^ins.  |^,  -W-,  H- 

6.  |  of  f  ,  f  of  |,  i  of  |  of  f  of  2f  .     Ant.  ffft,  |f*»  iff- 

EEMARK.  —  When  the  terms  of  the  fractions  are  small,  and  one 
denominator  is  a  multiple  of  the  others,  reduce  the  fractions  to  a 
common  denominator,  by  multiplying  both  terms  of  each  by  such  a 
number  as  will  render  its  denominator  the  same  as  the  largest  de- 
nominator. This  number  will  be  found  by  dividing  the  largest  denomina- 
tor by  the  denominator  of  the  fraction  to  be  reduced. 

PROBLEM.  —  Reduce  J  and  f  to  a  common  denominator. 

OPERATION. 

SOLUTION.  —  The   largest    denominator,    6,   is    a  1X2       2 

multiple   of     3;    therefore,    if   we   multiply   both 


terms    of    J  by  6  divided    by  3,  which  is  2,  it  is  5 

reduced    to  ^.  "7T         ==~r' 


84  RA  Y'S  HI  GHEE  ARITHMETIC. 

Reduce  to  a  common  denominator: 

1.  i,  |,  and  f.  Am.  f,  f,  |. 

2.  |,  |  and  TV  -4ns.  T82,  jf ,  -fa. 

3.  |,  f ,  ^  and  ^.  4n«.  |f ,  |£,  £f,  ft. 

CASE   VII. 

• 

139.  To  reduce  fractions  of  different  denominators, 
to  equivalent  fractions  having  the  least  common  de- 
nominator. 

PROBLEM. — Keduce  -f,  f ,  and  -fa  to  equivalent  fractions, 
having  the  least  common  denominator. 

OPEllATION. 

SOLUTION. — Find  the  L.  C.  M.  of  8,  9,  5      5X9      45 

and  24,  which  is  72 ;  divide  72  by  the  given  "7  —  g~\7a  ~~7~2 

denominators   8,  9,  and  24,  respectively ;  7      7  V  8      5  6 

multiply  both  terms  of  each  fraction  by  "—            =  — 

the  quotient  obtained  by  dividing  72  by  q                   _ 

its  denominator ;   the  L.  C.  M.  is  the  de-  —  = =  — 

nominator  of  the  equivalent  fractions.     In 

practice  it  is  not  necessary  to   multiply  each  denominator  in  form. 

Rule. — 1.  Find  the  L.  C.  M.  of  the  denominators  of  the 
given  fractions,  for  the  L.  C.  D. 

2.  Divide  this  L.  C.  D.  by  the  denominator  of  each  fraction, 
and  multiply  the  nwnerator  by  the  quotient. 

3.  Write  the  product  after   each  multiplication  as  a  numer- 
ator above  the  L.  C.  D. 

REMARK.— All  expressions  should  be  in  the  simplest  form. 

Reduce  to  the  least  common  denominator : 

1.  i,  \,  f  Am.  &,  &,  ||. 

2-  i,  f,  T%,  f-  Am.  ft,  if,  if,  M- 

3-  f,  I,  -H>  An*.  U,  n,  U- 

4-  |,  |,  -h,  M-  Ans.  f,  f,  1,  f. 

5-  f,  A.  H,  A-  -Ans.   !£,  fi  M,  ft- 
6.  If,  3|,  and  ^  of  3f  .Atw.  W>  W.  If- 


ADDITION  OF  FRACTIONS.  85 


ADDITION  OF  FKACTIONS. 

140.  Addition  of  Fractions  is  the  process  of  uniting 
two  or  more  fractional  numbers  in  one  sum. 

REMARK.  —  As  integers  to  be  added  must  express  like  units  (Art. 
52),  so  fractions  to  be  added  must  express  like  parts  of  like  units. 

PROBLEM.  —  What  is  the  sum  of  f,  |,  and  ^-? 

SOLUTION.  —  Reducing  the  given  fractions  to  OPERATION. 

equivalent  fractions  having  a  common  denom-  J=  J|        |  —  Jf 

inator,  we   have  -|-f,  |f,  and  JJ.     Since  these  T72  =  i| 

are  now  of  the  same  kind,  they  can  be  added  if  +  M  +  if  —  f  f 

by  adding  their   numerators.     Their   sum  is  fi  —  lfi>  Am. 

11=  iff- 

Rule.  —  Reduce  the  fractions  to  a  common  denominator,  add 
their  numerators,  and  write  the  sum  over  the  common  denom- 
inator. 

REMARKS.  —  1.  Each  fractional  expression  should  be  in  its  sim- 
plest form  before  applying  the  rule. 

2.  Mixed  numbers   and  fractions  may  be  added  separately  and 
their  sums  united. 

3.  After  adding,  reduce  the  sum  to  its  lowest  terms. 

EXAMPLES  FOR  PRACTICE. 

1.  |,  f  ,  and  ^.  Ans.  ff, 

2.  |,  |,  f  ,  and  ^.  Ans.  2J£ 

3.  If  and  2f  .  Ans. 

4.  2i    3f  and  4f  Ans. 


6.  li,  2|,  3£,  and  44, 

7.  |  of  f  ,  and  f  of  |  of 


9-  1  +  H  +  «  +  II  +  If  ^s.  4T\V 

10.  f  of  96^  +  f  of  ||  of  5f 

11-  1  +  1  +  If  +  lf  +  111  +  HI- 


86  RAY'S  HIGHER  ARITHMETIC. 


SUBTKACTION  OF  FRACTIONS. 

141.  Subtraction  of  Fractions  is  the  process  of  finding 
the  difference  between  two  fractional  numbers. 

REMARK. — In  subtraction  of  integers,  the  numbers  must  be  of 
Like  units  (Art.  54) ;  in  subtraction  of  fractions,  the  minuend  and 
subtrahend  must  express  like  parts  of  like  units. 

PROBLEM. — Find  the  difference  between  f  and  -f%. 

SOLUTION. — Reducing  the  given  frac-  OPERATION. 

tions  to  equivalent  fractions  having  a  f  ==f& 

common  denominator,  we  have  f  =  f  J,  T8^  =  ^f 
and  A  =  t$;   the  difference  is  &  =  A- 


Rule. — Reduce  the  fractions  to  a  common  denominator,  and 
write  the  difference  of  their  numerators  over  the  common  denom- 
inator. 

REMARK. — Before  applying  the  rule,  the  fractions  should  be  in 
their  simpl  st  form.  The  difference  should  be  reduced  to  its  lowest 
terms. 

EXAMPLES  FOR  PRACTICE. 

1.  f  —A.  Ans.  |f. 

2.  T8T  —  -ft*-  of  ^.  Ans. 

4.  ^-  —  y1^-  of  4.  Ans. 

5.  ii  — -gV  Am.  f|. 

f*         9  1 

REMARK. — When  the  mixed  numbers  are  small,  reduce  them  to 
improper  fractions  before  subtracting;  if  they  are  not  small,  sub- 
tract whole  numbers  and  fractions  separately,  and  then  unite  the 
results.  Thus, 


MULTIPLICATION  OF  FRACTIONS.  87 

PROBLEM.— Subtract  23 jf  from  31f. 

SOLUTION. — Reducing  the  fractions  to  a  OPERATION. 

common  denominator,  £  =  if.  But  if-  can  3 1  £  j^f  +  if  —  ft 
not  be  taken  from  if.  Take  a  unit,  if,  23j|  ft  — if  — it 
from  the  integer  of  the  minuend  and  add  it  7  .u. 

to  if.     Then  if  +  if  =  ff,  and  «  —  4*  = 
f|.     30  —  23  —  7.     Therefore  the  answer  is  ?if . 


9.  12f-10ff. 
10.  12||  —  9«. 

Z  o  O  O 

nR23          O2 
•    °  3T  —  *f ' 
-^2     7_J> Ql 

13.  15 —  f 

14.  18  — 5f.  Ans.  12|. 

15.  |  of  2%  —  3f|.  Jlns.  ff. 

16.  31— f  of  If.  4n«.  If. 

17.  -1/  qf  41  —  -^  of  3£  =*=  what?  Ans.  13f . 

18.  11-|  -f  8-J-  —  9^f  =  what  ?  .4ns.  10^|. 

19.  A  man  owned  f  f  of  a  ship,  and  sold  f  of  his  share : 
how  much  had  he  left?  Ans.  -f$. 

20.  After  selling  ^  of  f  +  |  of  f  of  a  farm,  what  part 
of  it  remains  ?  Ans.  •§--§-. 

21.  3i  +  4|  --  5£  +  16|  --  7Ji  +  10  —  14f,  is  equal 
to  what?  Ans.  6|f. 

22.  5J-  —  2|  +  -^  —  3f^  +  3^  +  8-J-  —  16|,  is  equal 
to  what?  Ans.  -|-|. 

23.  1—  f  of  f  —  |  of  f  =  what?  ^s.  ^. 

MULTIPLICATION  OF  FEACTIONS. 

142.  Multiplication  of  Fractions  is  finding  the  product 
when  either  or  when  each  of  the  factors  is  a  fractional  num- 
ber. There  are  three  cases : 

1.  To  multiply  a  fraction  by  an  integer. 

2.  To  multiply  an  integer  by  a  fraction. 

3.  To  multiply  one  fraction  by  another. 


88  RAY' S  HIGHER  ARITHMETIC. 

NOTE. — Since   any  whole  number  may  be  expressed  in  the  form 
of  a  fraction,  the  first  and  second  cases  are  special  cases  of  the  third. 

PROBLEM. — Multiply  f  by  -f. 

SOLUTION. — Once   f  is   f .     J  times  f  is  |-  of  OPERATION. 

f  =  &-    I   times  ?>  then>  is  5  times  s3o=ii-        f  X  f  =  i{!-, 

PROBLEM. — Multiply  f  by  6. 

SOLUTION. — Six  times  3  fourths  is  18  OPERATION. 

fourths.    Reducing  to  its  simplest  form,  J  X  f  —  ~V~  ~  4  i> 

we  have  4|.  or,  f  X  6  =  \8-  =  4  J 

PROBLEM. — Multiply  8  by  f. 

SOLUTION. — One  fifth  times  8  is  f ,  3  OPERATION. 

filths   times  8  is  3  times  f=2/-.     Ee-  f-  X  f  —  *£  =  4J,  Ans. 

ducing  to  a  mixed  number,  we  have  4f.  or, 


NOTE. — The  three  operations  are  alike,  and  from  them  we  may 
derive  the  rule. 

Rule. — Multiply  the  numerators  together  for  the  numerator 
of  the  product,  and  the  denominators  for  the  denominator  of  the 
product 

REMARK. — Whole  numbers  may  be  expressed  in  the  form  of 
fractions. 

EXAMPLES  FOR  PRACTICE. 

4.  TVX28.  Am.  15}. 

5.  if  X  30.  Ans.  26. 

6.  3f  X  5.  Ans. 


1.  if  X  12.         Ans. 

2.  -14  X  18.  Ans.  8-k 


, 


3.  ffX24.        Ans.  HI. 


OPERATIONS. 

REMARK. — In  multiplying  a  0  2  o  2 n 

mixed  number  by  a  whole  mini-  ^  °  1 1.  vx  5  ¥  _  5^5 

ber,  multiply  the  whole  number        -T-T— 

,,     r      %  15  ^/  =  18i 

and  the  traction  separately,  and  j 

add   the    products ;    or,  reduce        

the  mixed  number  to   an   im-  ¥>       s' 

proper  fraction,  and  multiply  it ;  as,  in  the  last  example. 


MULTIPLICATION  OF  FRACTIONS  89 


t  102f. 
Am.  A. 


7.  45  X  f 

Am.  35. 

11.  28X3}. 

8.  50XH- 

Ans.  39f 

12-  if  X  T^- 

9.  25  Xf. 

^1?18.    18|. 

13.  ifX-ft- 

10.  32X2|. 

Ans.  76. 

14-  If  x|f- 

15.  What  will  3-|-  yards  of  cloth  cost,  at  $4-J-  per  yard? 

BEMARK. — In  finding  the  product  of  two  mixed  numbers,  it  is 
generally  best  to  reduce  them  to  improper  fractions  ;  thus, 


OPERATION. 

SOIAJTION. — The  operation  may  be  performed  without  $  4  J 

reducing  to  improper  fractions ;  thus,  3  yards  will  cost  3  j- 

$13 J,  and  J  of  a  yard  will  cost  J  of  $4J  =  $1J;  hence,  1  3  \ 

the  whole  will  cost  $15.  1  i. 

Ans,  $15 

16.  6|  X  4J.        4w8.  30.       18.  12f  X  3T3T.       ^?is.  40J. 

17.  4|  X  2|.       Ans.  124.       19.  7|£  X  3^.       Ans.  26|. 

20.  Multiply  |  of  8  by  \  of  10.  Ans.  4. 

21.  Multiply  |  of  5|  by  f  of  3£.  Ans.  7£. 

22.  Multiply  f  of  |  of  5f  by  f  of  3|.  Ans.  4^. 

23.  Multiply  5,  4J,  2^,  and  f  of  4f .  ^?is.  94f. 

24.  Multiply  -f,  f ,  T5T,  |  of  2£,  and  -f-  of  3|.         ulns.  ^\%. 

25.  Multiply  f ,  £,  ^-,  31,  and  3|.  Jns.  4. 

26.  Multiply  3£,  4|,  5|,  |  of  T5T^and  6f .  ^Lns.  49. 

27.  At  %  of  a  dollar  per  yard,   what   will  25   yards  of 
cloth  cost?  $21|. 

28.  A  quantity  of  provisions  will  last  25  men  12f  days: 
how  long  will  the  same  last  one  man  ?  318|  days. 

29.  At   3|-  cents   a    yard,  what  will   2|    yards   of   tape 
cost?  9f  cents. 

30.  What   must  be   paid   for  f  of  f  of  a  lot  of  groceries 
that  cost  $18f  ?  87|. 

31.  K  owns  f  of  a  ship,  and   sells  f  of  his  share  to  L: 

what  part  has  he  left?  A. 

H.  A.  8. 


90  RAY'S  HIGHER  ARITHMETIC. 


DIVISION  OF  FKACTIONS. 

143.  Division  of  Fractions  is  finding  the  quotient 
when  the  dividend  or  divisor  is  fractional,  or  when  both 
are  fractional.  There  are  three  cases: 

1.  To  divide  a  fraction  by  an  integer. 

2.  To  divide  an  integer  by  a  fraction. 

3.  To  divide  one  fraction  by  another. 

NOTE. — Since  any  whole  number  may  be  expressed  in  the  form  of 
a  fraction,  the  first  and  second  cases  reduce  to  the  third  case. 

PROBLEM.— Divide  f  by  f. 

SOLUTION. — \    is  contained    in   1,  seven  OPERATION. 

times ;  \  is  contained  in  -J,  £  of  7  =  |  times ;  ^-r-|  =  fXi  =  ff 
\  is  contained  in  f ,  5  X  J  —  V  times ;  ^  is 

contained  in  f ,  J  of  -%5-  =  f  J  times.  It  will  be  seen  that  the  terms  of 
the  dividend  have  been  multiplied,  and  that  the  terms  of  the  divisor 
have  exchanged  places.  Writing  the  terms  thus  is  called  "  inverting 
the  terms  of  the  divisor,"  or  simply,  "  inverting  the  divisor." 

PROBLEM. — Divide  3  by  |. 

OPERATION. 

SOLUTION.— |  is  contained  in  f  -f-  f  —  f  X  f  =  V  =  7  J,  Ans. 
1,  five  times;  J  is  contained  in  Or,  3  H-  f  =  *£-==  7  J. 

3,  three  times  5  times  =  15  times ; 
f  is  contained  in  3,  J  of  15  times  =*£-  =  7±  times. 

PROBLEM. — Divide  f  by  2. 

OPERATION. 

SOLUTION.— Two  is  contained  f  ^-  f  =  f  X  i  =  T\  =  f >  ^r'*- 
in  1,  J  times;  2  is  contained  in  Or,  f  -5-  2  =  f. 

} ,  |  of  }  =  ^  times  ;  2   is  con- 
tained in  f ,  4  times  j1^  =  T4¥  =  -f-  times. 

NOTE. — From  these  solutions  we  may  derive  the  following  rule. 

Rule. — Multiply  the  dividend  by  the  divisor  with  its  term* 
inverted. 


DIVISION  OF  FRACTIONS. 


91 


REMARK. — The  terms  of  the  divisor  are  inverted  because  the 
solution  requires  it.  The  same  may  be  shown  by  a  different  solu- 
tion, as  below. 

PROBLEM.— Divide  f  by  f. 

SOLUTION. — Reduce  both  dividend   and  OPERATION. 

divisor  to   a  common    denominator.     The  f  =  J§      f  =  f  J  • 

quotient  of  Jf  -*-  f  |  is  the  same  as  10  -H  27          Jg  -r-  f  |  =  i£,  ^4ns. 
=  ^f.    The   same   result   is    obtained    by 

multiplying  the  dividend  by  the  divisor,  with  its  terms  inverted ; 
thus,  f  X  f  =  i?. 

REMARK. — Mixed  numbers  must  be  reduced  to  improper  frac- 
tions. Use  cancellation  when  applicable. 


EXAMPLES  FOR  PRACTICE. 


Ans. 
Ans. 
Ans. 

~9. 
.  23*. 


4V 


10.  If  — 5. 

11.  s*  —  f. 

12.  19*-*-!*, 

13.  73|-|-9f 
14. 


i-JH-3. 

2.  **-»7. 

3.  |--8. 
4>  6  *    2^ 

5.  21-=--ft, 

6.  £-=-*. 


15.  Divide  1*  by  *  of  f  of  7*. 

16.  Divide  T%-  of  T3¥  of  ^  by  ^  of  *|. 

17.  Divide  |  of  *  by  *  of  f. 

18.  Divide-*  of  3f  by  ||  of  7. 

19.  Divide  *  of  f  X  •£&  by  ^  of  3*. 

20.  Divide  f  of  5*  by  |-  of  ^  of  3*. 

21.  Divide  *  of  *  of  &  by  |  of  *  of  f 

O  /  11»/O  O  / 

22.  Divide  If  times  4|  by  1^  times  3|. 

23.  Divide  3f  by  f  of  8*  times  T5T  of 

24*  Divide  T9T  of  f  of  27*  by  |  of  T3T  of  5*. 
25.  What  is  2|  X  f  of  19*  -T-  (4f  X  f\  of  8) 


Ans. 
Ans. 


Am.  121. 


.  lOf. 


Am. 


Am. 


if 

A- 
if- 


? 


92  RAY'S  HIGHER  ARITHMETIC. 

144.     To  reduce  complex  to  simple  fractions. 
PROBLEM. — Reduce  -—  to  a  simple  fraction. 

OPERATION. 

EXPLANATION. — The  mixed        If  =  -y-  2J-  =  f 

numbers   are  reduced    to    im-        ±£-  -5-  |  =  -1g1-  X  I  =  if-  =  ff>  ^?'s- 
proper     fractions,      and     the 
numerator  is  divided   by  the  denominator.     (Art.  114.) 

Rule. — Divide    the    numerator  by   the    denominator,  as    in 
division  of  fractions. 

Reduce  to  simple  fractions: 


1. 


2. 


-. 


3.    A 


Ans.  -£%. 

Am.  f. 

Ans.  If. 


4.         . 


6. 


Ans.  f  J-. 


16H' 


REMARK. — Complex  fractions  may  be  multiplied  or  divided,  by 
reducing  them  to  simple  fractions.  The  operation  may  often  be 
shortened  by  cancellation. 


7.         X      . 


8-      X 

9-    x 


10-!-T2T    Ans- 


40$       73 ' 


Am.  U. 

o  o 

Ans.  3 


THE  GREATEST  COMMON  DIVISOR  OF  FRACTIONS. 

145.  The  greatest  common  divisor  of  two  or  more 
fractions  is  the  greatest  fraction  that  will  exactly  (Jivide 
each  of  them. 


G.  C.  D.  OF  FRACTIONS.  93 

One  fraction  is  divisible  by  another  when  th'e  numerator 
of  the  divisor  is  a  factor  of  the  numerator  of  the  dividend, 
and  the  denominator  of  the  divisor  is  a  multiple  of  the 
denominator  of  the  dividend. 

Thus,  &  is  divisible  by  &  ;  for  T\  =  if;  ff  Hh  A  =  6- 

The  greatest  common  divisor  of  two  or  more  fractions, 
must  be  that  fraction  whose  numerator  is  the  G.  C.  D.  of 
all  the  numerators,  and  whose  denominator  is  the  L.  C.  M. 
of  the  denominators. 

Thus,  the  G.  C.  D.  of  &  and  J|  is  7f^. 

PROBLEM. — Find  the  G.  C.  D.  of  -|,  f -|,  and  -fy. 

SOLUTION. — Since  5  and  7  are  both  OPERATION. 

prime  numbers,  1  is  the  G.  C.  D.  of  1  =  G.  C.  D.  of  5,  25,  7 

all  the  numerators  ;  96  is  the  L.  C.  9  6  =  L.  C.  M.  of  8,  32,  12 

M.  of  8,  32,  and  12;  therefore,  the  G.  ^,  Ans. 
C.  D.  of  the  fraction  is  -g1^. 

Rule. — Find  the  G.  G.  D.  of  the  numerators  of  the  fractions, 
and  divide  it  by  the  L.  C.  M.  of  their  denominators. 

REMARK. — The  fractions  should  be  in  their  simplest  forms  before 
the  rule  is  applied. 

Find  the  greatest  common  divisor: 

1.  Of  83i  and  268f.  Ans.  2TV 

2.  Of  14TV  and  95f.  Am.  &. 

3.  Of  591  and  735-}-f.  Ans.  2|f. 

4.  Of  23TV  and  213if.  Ana.  2J|. 

5.  Of  418|  and  17721.  £ns.  jf. 

6.  Of  261  jf  and  652^.  Ans.  4|f. 

7.  Of  44|,  546|,  and  3160.  Ans.  4|. 

8.  A  farmer  sells  137^  bushels  of  yellow  corn,  478^  bushels 
of  white  corn,  and  2093f  bushels  of  mixed  corn :  required 
the  size  of  the  largest  sacks  that  can  be  used  in  shipping,  so 
as  to  keep  the  corn  from  being  mixed ;  also  the  number  of 
sacks  for  each  kind.  3£  bushels ;  44,  153,  and  670. 


94  RAY'S  HIGHER  ARITHMETIC. 

9.  A  owns  a  tract  of  land,  the  sides  of  which  are  134f , 
128|-,  and  115^  feet  long:  how  many  rails  of  the  greatest 
length  possible  will  be  needed  to  fence  it  in  straight  lines, 
the  fence  to  be  6  rails  high,  and  the  rails  to  lap  6  inches  at 
each  end?  354  rails. 


THE  LEAST  COMMON  MULTIPLE  OF  FRACTIONS. 

146.  The  least  common  multiple  of  two  or  more 
fractions  is  the  least  number  that  each  of  them  will  divide 
exactly. 

NOTE. — The  G.  C.  D.  of  several  fractions  must  be  a  fraction,  but 
the  L.  C.  M.  of  several  fractions  may  be  an  integer  or  a  fraction. 

A  fraction  is  a  multiple  of  a  given  fraction  when  its 
numerator  is  a  multiple,  and  its  denominator  is  a  divisor,  of 
the  corresponding  terms  of  the  given  fraction. 

ILLUSTRATION. — T8r  is  a  multiple  of  -fa.  8  is  a  multiple  of  2,  and 
11  is  a  divisor  of  33;  hence,  T8T  -r-  -fa  =  T8T  X  ¥  =  I2-  The  same 
result  is  otherwise  obtained ;  thus,  T8T  =  f  f ,  and  f f  -5-  -fa  =  12. 

A  fraction  is  a  common  multiple  of  two  or  more  given 
fractions  when  its  numerator  is  a  common  multiple  of  the 
numerators  of  the  given  fractions,  and  its  denominator  is  a 
common  divisor  of  the  denominators  of  the  given  fractions. 

A  fraction  is  the  least  common  multiple  of  two  or  more 
fractions  when  its  numerator  is  the  least  common  multiple 
of  the  given  numerators,  and  its  denominator  is  the  greatest 
common  divisor  of  the  given  denominators. 

PROBLEM. — Find  the  L.  C.  M.  of  -^,  f ,  and  -f. 

SOLUTION. — The  L.  C.  M.  of  OPERATION. 

the  numerators  is  15.     The  G.  L.  C.  M.  of  1,  3,  5  =  3  X  5  =  1  5 

C.  D.  of  the  denominators  is  1 ;  G.  C.  D.  of  3,  4,  6  =  1 

therefore,  the  L.  C.  M.  of  the  .*.  y-,  Ans. 
fractions  is  ^,  or  15. 


L.  C.  M.  OF  FRACTIONS.  95 

Rule. — Divide  the  L.  C.  M.  of  the  numerators  by  the  G.  C. 
D.  of  the  denominators. 

KEMABK. — The  fractions  must  be  in  their  simplest  forms  before 
the  rule  is  applied. 

Find  the  least  common  multiple : 

1-  Of  f  >  t>  f>  f>  and  f  An*.  60. 

2.  Of  4£,  6f ,  5f ,  and  10J.  Ans.  4721, 

3.  Of  31,  4f ,  -3^-,  5f ,  and  12£.  4wa.  350. 

4.  A  can  walk  around  an  island  in  14^  hours ;  B,  in  9^j 
hours ;  C,  in  16f  hours  ;  and  D,  in  25  hours.     If  they  start 
from  the  same  point,  and  at  the  same  time,  how  many  hours 
after  starting  till  they  are  all  together  again  ?         100  hours. 


PROMISCUOUS   EXERCISES. 

NOTE  TO  TEACHERS. — All  problems  marked  thus  [*],  are  to  be 
solved  mentally  by  the  class.  In  the  solution  of  such  problems,  the 
following  is  earnestly  recommended: 

1.  The  teacher  will  read  the  problem  slowly  and  distinctly,  and 
not  repeat  it. 

2.  The  pupil  designated  by  the  teacher,  will  then  give  the  answer 
to  the  question. 

3.  Some  pupil,  or  pupils,  will  now  reproduce  the  question  in  the 
exact  language  in  which  it  was  first  given  to  the  class. 

4.  The  pupil,  or  pupils,  called  upon  by  the  teacher,  will  give  a 
short,  logical  analysis  of  the  problem. 

1.  What  is  the  sum  of  3|,  4J-,  5J,  f  of  |,  and  £  of  £ 
of  |?  13f|. 

2.  The  sum  of  l^-  and  —  is  equal  to  how  many  times 
their  difference?  9"  5  times. 

3.  What  is  (2* +  5  of  —  _M)-=-lJLL?  5. 

2       3        27 


96  &A  Y>  S  HIGHER  ARITHMETIC. 


4.  Reduce  lM    and    I  X  (100  -  ^  +  Ii)    to 

5i  -  4J-  7  3      h  2i; 

their  simplest  forms.  16  and  26T2^. 

5.  What  is  ±  of  5J  —  -fr  of  3£?  ^V 

6.  What  is  |f  X  TT52  X  iff  X  If  equal  to?  TV 

7*   1  vJ—  nivAni?  ,1 

~2~X~I~X~~3~ 

g      1       l-i       2-i      3-|      4-|? 
-X--X        -  >— 


(2-i)X(4-3f) 

10.  jj  X  4i  X  4^-  —  1  ^  what?  42^ 

11.  Add  |  of  f  of  |,  |  X  |  of  1 J,  and  £.  |f 

12.  f  of  -y~  of  what   number,   diminished    by    — H — , 

leaves  -|-|?  y9^-. 

13.*  James's  money  equals  f  of  Charles's  money ;  and  £ 
of  James's  money  -f-  $33  equals  Charles's  money :  how  much 
has  each?  James,  $36;  Charles,  $60. 

14.*  A  leaves  L  for  N  at  the  same  time  that  B  leaves  N 
for  L.  The  two  places  are  exactly  109  miles  apart :  A 
travels  1\  miles  per  hour,  and  B,  8J  miles  per  hour;  in 
how  many  hours  will  they  meet,  and  how  far  will  each  have 
traveled?  6ff  hours.  A,  5 Iff-  miles;  B,  57-22T  miles. 

15.  What    number    multiplied    by  •§•  of  f  of  3-fJ-    will 
produce  2J?  2|f. 

16.  What,  divided  by  If,  gives  14f?  23|. 

17.  What,  added  to  14f,  gives  29|f  ?  15^. 
18.*  I  spend  f  of  my  income  in  board,  \  of  it  in  clothes, 

and  save  $60  a  year:  what  is  my  income?  $216. 

19.*  Divide  51  into  two  such  parts  that  -|  of  the  first  is 

equal  to  f  of  the  second.  27  and  24. 

20.*  \  is  what  part  of  |?  f. 


COMMON  FRA  CTIONS.  97 

21.  Divide  $  of  3f  by  |f  of  7 ;  and  ^  of  |  of  27£  by  f 
of  T3T  of  5±.  |f  and  34|f 

22.  Multiply  T7T  of  2£  by  •&  of  19£;  and  divide  $  of  f 
of  14}  by  T8T  of  f  of  13f  7^  and  &&. 

23.  (Mil  +  iffl  +  tttt)  -  f  =  what?  5f. 

24.*  A  bequeathed  -^  of  his  estate  to  his  elder  son;  the 
rest  to  his  younger,  who  received  $525  less  than  his  brother. 
What  was  the  estate?  $5250. 

25.  Find  the  sum,  difference,  and  product  of  3£  and  2^ ; 
also,  the  quotient  of  their  sum  by  the  difference. 

Sum  5f|,  diff.  Iff,  prod.  8}f ;    quot.  3^°7. 

26.*  A  cargo  is  worth  7  times  the  ship :  what  part  of  the 
cargo  is  T5g-  of  the  ship  and  cargo?  -f^. 

27.  By  what  must  the  sum  of  ^7T>  AW  and  ATT  be 
multiplied  to  produce  1000?  1000. 

28.  Multiply  the  sum  of  all  the  divisors  of  8128,  includ- 
ing 1,  by  the  number  of  its  prime  factors  excluding  1,  and 
divide  by  14%  381. 

29.  6|-  is  what  part  of  10T7T?     Keduce  to  its  simplest 
formf-!  +  f--li. 

30.  Multiply  1,  14f,  ?i,    I,    g,  and  6. 

31.  |  of  |  of  what  number  equals  9f|?  20. 
32.*  A  63-gallon  cask  is  f  full :  9|  gallons  being  drawn 

off,  how  full  will  it  be?  |||. 

33.  If  a  person    going   3f   miles    per    hour,   perform  a 
journey  in  14f  hours,  how  long  would  he  be,  if  he  traveled 
5^  miles  per  hour?  lO^f  hours. 

34.  A   man   buys  32f  pounds  of  coffee,   at  17f  cents  a 
pound:  if  he  had  got  it  4|  cents  a  pound  cheaper,  how  many 
more  pounds  would  he  have  received?  UTTI  P°unds. 

35.  Henry  spent  ^  of  his  money  and  then  received  $65 ; 
he  then  lost  f  of  all  his  money,  and  had  in  hand  $10  less 
than  at  first.     How  much  had  he  at  first?  $33. 

H.  A.  9. 


98 


HAY'S  HIGHER  ARITHMETIC. 


Topical  Outline. 


COMMON  FRACTIONS. 


1.  Definition. 


2.  Classes.. 


8.  Terms... 


1.  As  to  Kinds.... 


2.  As  to  Value.. 


3.  As  to  Form.... 


1.  Numerator.. 

2.  Denominator. 

3.  Similar. 

.  4.  Dissimilar. 


{1.  Common. 
2.  Decimal. 
'  1.  Proper. 
Improper. 
Mixed. 
Simple. 
Complex. 
1 3.  Compound. 


f  1.  Pro 
•j  2.  Imi 
I  3.  Mb 
fl.  Sir 
\  2.  Coi 
1 3.  Co: 


4.  Principles. 


5.  Reduction...  - 


1.  Cases 


2.  Principles. 


3.  Rules. 


1.  Lowest  Terms. 

2.  Higher  Terms. 

3.  Mixed  Numbers  to  Improper  Fractions. 

4.  Improper  Fractions  to  Mixed  Numbers. 

5.  Compound  to  Simple  Fractions. 

6.  Common  Denominator. 

7.  Least  Common  Denominator. 


6.  Practical  Applications 


f 

fl.  Definition. 

1.  Addition  

-j  2.  Principles. 

1  3.  Rule. 

fl.  Definition. 

2.  Subtraction  

-j  2.  Principles. 

I  3.  Rule. 

fl.  Definition. 

3.  Multiplication  

-j  2.  Principles. 

I  3.  Rule. 

fl.  Definition. 

4.  Division  

-j  2.  Principles. 

[  3.  Rule. 

5.  Divisors,  Multiples,  etc. 


IX.  DECIMAL  FRACTIONS. 

147.  A  Decimal  Fraction  is  a  fraction  whose  denom< 
inator  is  10,  or  some  product  of  10,  expressed  by  1  with 
ciphers  annexed. 

REMARK  1. — A  decimal  fraction  is  also  defined  as  a,  fraction  whose 
denominator  is  some  power  of  10.  By  the  "power"  of  a  quantity,  is 
usually  understood,  either  that  quantity  itself,  or  the  product 
arising  from  taking  only  that  quantity  a  certain  number  of  times 
as  a  factor.  Thus,  9  =  3  X  3,  or  the  second  power  of  3. 

KEMARK  2.— Since  decimal  fractions  form  only  one  of  the  classes 
(Art.  108)  under  the  term  fractions^  the  general  principles  relat- 
ing to  common  fractions  relate  also  to  decimals. 

148.  The  orders  of  integers  decrease  from  left  to  right 
in  a  tenfold  ratio  (Art.  48).     The  orders  may  be  continued 
from  the  place  of  units  toward  the  right  by  the  same  law 
of  decrease. 

149.  The  places  at  the  right  of  units  are  called  decimal 
places,  and  decimal  fractions  when  so  written,  without  a 
denominator  expressed,  are  called  decimals. 

150.  The  decimal  point,  or  separatrix,  is  a  dot  [ .  ] 
placed  at   the   left   of  decimals   to   distinguish   them    from 

integers. 

Thus,  TV        is  written        .1 

T*5         "  "  -01 

"        "         -001 
"        "       -0001 
From  this,  it  is  evident  that, 

The  denominator  of  any  decimal  is  1  with  as  many  ciphers 
annexed  as  there  are  places  in  the  decimal. 

151.  A  pure  decimal  consists  of  decimal  places  only ; 
as,  .325 

(99) 


100  JRAY'S  HIGHER  ARITHMETIC. 

152.  A  mixed  decimal  consists  of  a  whole  number  and 
a  decimal  written  together;  as,  3.25 

REMARK. — A  mixed  decimal  may  be  read  as  an  improper  frac- 
tion, since  3T2o^  =  ff£. 

153.  A  complex  decimal  has  a  common  fraction  in  its 
right-hand  place ;  as,  .033^ 

154.  From   the   general  law   of  notation   (Arts.  48  and 
148)  may  be  derived  the  following  principles : 

PRINCIPLE  \.—If,  in  any  decimal,  the  point  be  moved  to 
the  right,  the  decimal  is  multiplied  by  10  as  often  as  the  point  is 
removed  one  place. 

ILLUSTRATION. — If,  in  the  decimal  .032,  we  move  the  point  one 
place  to  the  right,  we  have  .32.  The  first  has  three  decimal  places, 
and  represents  thousandths;  while  the  second  has  two  places,  and 
represents  hundredth*.  (Art.  129,  Prin.  i.) 

PRINCIPLE  II. — If,  in  any  decimal,  the  point  be  moved  to  the 
left,  the  decimal  is  divided  by  10  as  often  as  the  point  is  removed 
one  place. 

ILLUSTRATION. — If,  in  the  decimal  .35  we  move  the  point  one 
place  to  "the  left,  we  have  .035.  The  first  represents  hundredths; 
the  second,  thousandths,  while  the  numerator  is  not  changed.  (Art. 
129,  Prin.  n.) 

PRINCIPLE  III. — Decimal  ciphers  may  be  annexed  to,  or 
omitted  from,  the  right  of  any  number  without  altering  its  value. 

ILLUSTRATION.— .5  is  equal  to  .500;  for  ^£T™  ==  fooo-  The 
reverse  may  be  shown  in  the  same  way.  (Art.  129,  Prin.  in.) 

NUMERATION    AND   NOTATION   OF   DECIMALS. 

155.  Since  .6  =  ^5   .06=^;    and  .006  =  TTftnr,  any 
figure  expresses  tenths,   hundredths,  or  thousandths,  according 
as   it  is  in   the  1st,  2d,  or  3d  decimal  place;  hence,  these 
places  are  named  respectively  the  tenths',  the  hundredths',  the 


DECIMAL   FRACTIONS.  101 

thousandths9  place  ;  other  places  are  named  in  the  same  way, 
as  seen  in  the  following  table  : 

TABLE  OF  DECIMAL  ORDERS. 


" 


A 

A 

AH 
£ 

a 

1  "•?  «  s  a  rf 

s  *3  .2  5  •&  .3 

Islf  If 

5  T3  2  S  'O  o 

Ill 

A,  PJ  -H  A  s  ^3 

3  P  g  S  0  3 

H  W  ^  S  K  -W 

1st 

place 

.2 

. 

.  read 

2 

Tenths. 

2d 

tt 

.08 

•         •         . 

ti 

8 

Hundredths. 

3d 

It 

.00 

5 

.         . 

it 

5 

Thousandths. 

4th 

It 

.  00 

0 

7  .     . 

it 

7 

Ten-thousandths. 

5th 

ii 

.00 

0 

03      . 

1  1 

3 

Hundred-thousandths. 

6th 

(i 

.00 

0 

001  . 

ft 

1 

Millionth. 

7th 

« 

.00 

0 

0009 

(t 

9 

Ten-millionths. 

8th      "      .00000004.     "     4  Hundred-millionths. 
9th      "      .000000006     "     6  Billionths. 

NOTE. — The  names  of  the  decimal  orders  are  derived  from  the 
names  of  the  orders  of  whole  numbers.  The  table  may  therefore  be 
extended  to  trillionths,  quadrillionths,  etc. 

PROBLEM. — Read  the  decimal  .0325 

SOLUTION. — The  numerator  is  325;  the  denomination  is  ten- 
thousandths  since  there  are  four  decimal  places.  It  is  read  three 
hundred  and  twenty-five  ten-thousandths. 

Rule. — Read  the  number  expressed  in  the  decimal  places  as 
the  numerator,  give  it  the  denomination  expressed  by  the  right- 
hand  figure. 

EXAMPLES  TO  BE  READ. 

1.     .9  3.     .Of 

4.     .035 


102 


RAY'S  HIGHER  ARITHMETIC. 


5. 

.7200 

6. 

.5060 

7. 

1.008 

8. 
9. 
10. 

9.00^ 
105.0f 
.0003 

11. 

00.100 

12. 

180.010 

13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 


2030.0 

40.68031 

200.002 

.0900001 

61.010001 

31.0200703 

.000302501 

.03672113 


EXERCISES  IN  NOTATION. 

156.  The  numerator  is  written  as  a  simple  number;  the 
denomination  is  then  expressed  by  the  use  of  the  decimal 
point,  and,  if  necessary,  by  the  use  of  ciphers  in  vacant 
places. 

PROBLEM. — Write  eighty-three  thousand  and  one  billionths. 

EXPLANATION. — First   write    the    numerator,  OPERATION. 

83001.     If    the  point  were    placed  immediately         .000083001 
at  the   left  of  the  8,  the  denominator  would  be 
hundred-thousandths;  it  is  necessary  to  fill  four  places  with  ciphers 
so  that  the  final  figure  may  be  in  billionths'  place. 

Rule. —  Write  the  numerator  as  a  whole  number;  then  place 
the  decimal  point  so  that  the  right-hand  figure  shall  be  of  the 
same  name  as  the  decimal. 


EXAMPLES  TO  BE  WRITTEN. 


1.  Five  tenths. 

2.  Twenty -two  hundredths. 

3.  One  hundred  and  four  thou- 

sandths. 

4.  Two    units    and     one    hun- 

dredth. 


5.  One   thousand    six    hundred 

and  five  ten-thousandths. 

6.  Eighty-seven     hundred-thou- 
sandths. 

7.  Twenty-nine    and     one     half 

ten-millionths. 


REDUCTION  OF  DECIMALS. 


103 


8.  Nineteen    million   and    one 
billionths. 

9.  Seventy  thousand  and  forty- 

two  units  and  sixteen  hun- 
dredths. 

10.  Two     thousand    units     and 
fifty-six  and  one  third  mill- 
ionths. 

11.  Four  hundred   and    twenty- 
one  tenths. 

12.  Six  thousand  hundredths. 

13.  Forty-eight    thousand  three 
hundred     and    five     thou- 
sandths. 


14.  Eight  units  and  one  half  a 
hundredth. 

15.  Thirty-three     million     ten- 

millionths. 

16.  Four  hundred   thousandths. 

17.  Four    hundred-thousandths. 

18.  One    unit    and  one   half    a 
billionth. 

19.  Sixty -six  thousand  and  three 

millionths. 

20.  Sixty-six  million  and  three 
thousandths. 

21.  Thirty -four    and    one   third 

tenths. 


[REDUCTION  OF  DECIMALS. 

157.     Reduction  of  decimals  is    changing  their   form 
without  altering  their  value. 


CASE    I. 

158.     To  reduce  a  decimal  to  a  common  fraction. 
PROBLEM. — Reduce  .24  to  a  common  fraction. 


SOLUTION. — .24  is  equal  to  y2^,  which, 
reduced,  is  -fg. 


OPERATION. 

=  ^  =  A,  Am. 


PROBLEM. — Reduce  .12^  to  a  common  fraction. 

OPERATION. 

SOLUTION. — Write  12£  as  a  ^  2  i 

numerator,  and  under  it  place        .  1  2  J  =  — —  =^-^=  fifa  =  J,  Ans. 
100  as  a  denominator.      Re- 
duce   the    complex    fraction  according  to  Art.  144. 

Rule. —  Write  the   decimal  as  a  common  fraction;  then  re- 
duce the  fraction  to  its  lowest  terms. 


104 


KAY'S  HIGHER   ARITHMETIC. 


REMARK.— If  the  decimal  contains  many  decimal  places,  an 
approximate  value  is  sometimes  used.  For  example,  3.14159  — 
nearly  3^. 


Reduce  to  common  fractions . 


1. 

.25625 

Ans.  T4^jy. 

9. 

ll.Of 

2. 

.15234375 

Ans.  -£-£$• 

10. 

.390625 

3. 

2.125 

Ans.  2$. 

11. 

.1944| 

4. 

19.01750 

A  via   1  Q   7 
-<!'«*.  J-^nnT' 

12. 

.24|  . 

5. 

16.00^ 

,4ns.  16-j^-jr. 

13. 

.33| 

6. 

350.028^- 

Ans.  350^. 

14. 

.66| 

7. 

.6666661 

4ns.  |. 

15. 

.25 

8. 

.003125 

.4ns.  -g^. 

16. 

.75 

Ans.  lly-g-. 
Am.  f|. 
Ans.  fa 
Am.  li. 

Ans.  %. 

Am.  f. 

Am.  %. 

Am.  f . 


CASE    II. 


159.     To  reduce  common  fractions  to  decimals. 


OPERATION. 

8)7.000 

.  8  7  5,   Ans. 


PROBLEM. — Reduce  f-  to  a  decimal. 

SOLUTION.— The  fraction  |  =  J  of  7.  7  — 
7.0,  4-  of  7.0  =  .8,  with  6  tenths  remaining;  .6. 
=  .60,  J  of  .60^.07,  with  4  hundredths  re- 
maining; .04^.040,  J  of  .040^.005.  The  answer  is  .875. 

NOTE. — Another  form  of  solution  may  be  obtained  by  multiply- 
ing both  terms  of  the  fraction  by  1000;  dividing  both  terms  of  the 
resulting  fraction  by  the  first  denominator  and  writing  the  answer 
as  a  decimal.  Thus,  |  =  J{$$,  |$§g  =  T^V  =  .875.  Both  solutions 
depend  on  Art.  129,  Prin.  in. 

.  'Rule.— Annex  ciphers  to  the  numerator  and  divide  it  by  the 
denominator.  Then  point  off  as  many  decimal  places  in  the 
quotient  as  there  are  ciphers  annexed. 

REMARK. — Any  fraction  in  its  lowest  terms  having  in  its  de- 
nominator any  prime  factor  other  than  2  and  5,  can  not  be  reduced 
exactly  to  a  decimal.  Thus,  y12  =  .08333  +•  The  sign  +  is  used  at 


ADDITION  OF  DECIMALS.  105 

the  end  of  a  decimal  to  indicate  that  the  result  is  less  than  the  true 
quotient.  The  sign  —  is  also  sometimes  used  to  indicate  that  the 
last  figure  is  too  great.  Thus,  \  =  .1428  +,  or,  by  abbreviating,  £ 
=  .143—. 


Reduce 

to  decimals: 

1.  £. 

-4:718.    .75 

6.  £.                          Am.  .8 

2.  i 

Ans.  .125 

7.  Jft.                   J.918.  .495 

3.   2ir 

Jbis.  .05 

8.  /¥.              ^i§.  .078125 

4.  i-5-. 

^4?is.  .46875 

9.  TVV        ^w.  .05078125 

5-  Tinro 

Ans.  .005625 

10.  ^2  ¥•  ^  .0009765625 

NOTE.  —  The  rule  converts  a  mixed  number  into  a  mixed  decimal, 
and  a  complex  into  a  pure  decimal  ;  thus,  9|  =  9.375,  since  f  —  .375  ; 
and  .26/7  =  .2612,  since  &  =  .12. 


11.  16|-  Am.   16.5 

12.  42T\  Ans.  42.1875 

13.  .015i  Ans.  .01525 

14.  lOl.Olf  Am.  101.0175 

15.  751.19&  Ans.  75119.0375 

16.  2.00^  Am.  2.00003125 


ADDITION  OF  DECIMALS. 


160.     Addition  of  Decimals  is  finding  the  sum   of  two 
or  more  decimals. 

NOTE. — Complex  decimals,  if  there  are  any,  must  be  made  pure, 
as  far,  at  least,  as  the  decimal  places  extend  in  the  other  numbers. 

PROBLEM.— Add  23.8  and  17|  and  .0256  and  .41|. 

OPERATION. 

SOLUTION. — Write  the   numbers  23    8 

as  in  the  operation,  and  add  as  in  i  7  i  —  \  7  m  5 
simple   addition.     Write  the  deci-  .0256 

mal  point  in  the  sum  to  the  left  of  ^^  2—        .  4  ]  6  6  f 
tenths- 


106  RAY'S  HIGHER  ARITHMETIC. 

Rule. — 1.  Write  the  numbers  so  that  figures  of  the  same 
order  shall  stand  in  the  same  column. 

2.  Then  add  as  in  simple  numbers,  and  put  the  decimal  point 
to  the  left  of  tenths. 

REMARK. — The  proof  of  each  fundamental  operation  in  decimals 
is  the  same  as  in  simple  numbers. 

1.  Find  the  sum  of  1  +  .9475  1.9475 

2.  Of  1.331  added  to  2.66|  4. 

3.  Of  14.034,  25,  .000062^,  .0034  39.0374625 

4.  Of  83  thousandths,  2101  hundredths,  25  tenths,  and 
94^  units.  118.093 

5.  Of  .16|,  .37^,  5,  3.4|,  .000£  8.980^ 

6.  Of  4  units,  4  tenths,  4  hundredths.  4.44 

7.  Of  .11|  +  .6666f  +  .2222221  1. 

8.  Of  .14f,  .018f,  920,  .0139$,  920.1754 

9.  Of  16.008J,  .0074f,  .2f,  .00019042^  16.299768199f 

10.  Of  675  thousandths,  2  millionths,  64£,  3.489107,  and 
.00089407  68.29000307 

11.  Of  four  times  4.067£  and  .000^  16.272 

12.  Of  216.86301,  48.1057,  .029,  1.3,  1000.     1266.29771 

13.  Add    35   units,   35  tenths,    35   hundredths,  35   thou- 
sandths. 38.885 

14.  Add  ten  thousand  and  one  millionths ;  four  hundred- 
thousandths  ;  96  hundredths ;  forty-seven  million  sixty  thou- 
sand and  eight  billionths.  1.017101008 

SUBTRACTION  OF  DECIMALS. 

161.  Subtraction  of  Decimals  is  finding  the  difference 
between  two  decimals. 

PROBLEM. — From  6.8  subtract  2.057 

SOLUTION. — Write  the  numbers  so  that  units  of  OPERATION. 

the  same  order  stand  in  the  same  column ;  sup-  6  .  8 

pose  ciphers  to  be  annexed  to  the  8,  and  subtract  2.057 

as  in  whole  numbers.  4.743,  Ans. 


SUBTRACTION  OF  DECIMALS. 

PROBLEM.— From  13.256f  subtract  6.77£ 


107 


EXPLANATION. — In  this  example, 
the  complex  decimals  are  carried  out 
by  division  to  the  same  place,  and  the 
common  fractions  treated  by  Art.  141. 


OPERATION. 

13.256f 
6.771  =  6.773} 

6  .  4  8  3  f ,  Ans. 


Rule. — 1.  Write  the  subtraliend  beneath  the  minuend  so  tiiat 
units  of  tJie  same  order  stand  in  the  same  column. 

2.  Subtract  as  in  simple  numbers,  and  write  the  decimal 
point  as  in  addition  of  decimals. 

NOTES. — 1.  If  either  or  both  of  the  given  decimals  be  complex, 
proceed  as  directed  in  the  second  problem. 

2.  If  the  minuend  has  not  as  many  decimal  places  as  the  subtra- 
hend, annex  decimal  ciphers  to  it;  or  suppose  them  to  be  annexed, 
until  the  deficiency  is  supplied. 


EXAMPLES  FOR  PRACTICE. 


1.  Subtract  8.00717  from  19.54  11.53283 

2.  3  thousandths  from  3000.  2999.997 

3.  72.0001  from  72.01  .0099 

4.  Subtract  .93^  from  1.169f-  .238f 

5.  How  much  is  19  —  8.999^?  lO.OOOf 

6.  How  much  less  is  .04£  than  .4?  .35f 

7.  How  much  is  .65007  —  1?  .15007 

8.  What  is  2f  —  If  in  decimals  ?  .95 

9.  Subtract  1  from  1.684  .684 

10.  |  of  a  millionth  from  .OOOf  .000443|| 

11.  11  hundredths  from  49f  tenths.  4.9225 

12.  10000  thousandths  from  10  units.  0. 

13.  241  tenths  from  3701  thousandths.  1.251 

14.  1-|  units  from  1875  thousandths.  0.^ 

15.  f  of  a  hundredth  from  y1^  of  a  tenth.  0. 

16.  64£  hundredths  from  100  units.  99.35| 


108  RA  Y'S  HIGHER  ARITHMETIC. 


MULTIPLICATION   OF  DECIMALS. 

162.     Multiplication  of  Decimals  is  finding  the  product 
when  either  or  when  each  of  the  factors  is  a  decimal. 

PRINCIPLE. — The   number  of  decimal  places  in  the  product 
equals  the  number  of  decimal  places  in  both  factors. 

PROBLEM.— Multiply  2.56  by  .184 

OPERATION. 

SOLUTION.— 2.56  =  fj[*>     and    '  .184  =  ^ftfo.  2.56 

Now,  -f Jf  X TV8o4o  =  rWAV>  that  is,  the  product  .184 

of   Imndredths    by  thousandths    is    hundredth-  1024 

thousandths.       This     requires    five    places    of  2048 

decimals,   or   as   many  as   are   found    in   both  256 

'factors.  .47104,  Am. 

Rule. — 1.  Multiply  as  in  whole  numbers. 
2.  Point  off  as  many  decimal   places  in  the  product  as  there 
are  decimal  places  in  the  two  factors. 

REMARKS. — 1.  If  the  product  does  not  contain  as  many  places  as 
the  factors,  prefix  ciphers  till  it  does  contain  as  many. 

2.  Ciphers  to  the  right  of  the  product  are  omitted  after  pointing. 


EXAMPLES  FOR  PRACTICE. 

1.  1X.1  .1 

2.  16  X  .03^  .53£ 

3.  .OlX.li  .0015 

4.  ,080  X  80.  6.4 

5.  37.5  X82i  -3093.75 

6.  64.01  X. 32  20.4832 

7.  48000  X  73.  3504000. 

8.  64.66f  X  18.  1164. 

9.  .561  x. 03^  .0172-H- 
10.  738X120.4  88855.2 


MULTIPLICATION  OF  DECIMALS.  109 

11.  .0001  X  1.006  .0001006 

12.  34  units  X- 193-  6.562 

13.  27  tenths  X.4£  1.134 

14.  43. 7004  X- 008  '.3496032 

15.  21.0375  X  4.44£  93.5 

16.  9300.701  X  251.      .  2334475.951 

17.  430.0126X4000.  1720050.4 

18.  .059  X -059  X  .059  .000205379 

19.  42  units  X  42  tenths.  176,4 

20.  21  hundredths  X  600.  13.5 

21.  7100  X  i  of  a  millionth.  .0008875 

22.  26  millions  X  26  millionths.  676. 

23.  2700  hundredths  X  60  tenths.  162. 

24.  6.3029  X  .03275  .206419975 

25.  135.027  X  1.00327  135.46853829 

163.  Oughtred's  Method  for  abbreviating  multiplica- 
tion, may  be  used  when  the  product  of  two  decimals  is 
required  for  a  definite  number  of  decimal  places  less  than 
is  found  in  both  factors. 

PROBLEM.— Multiply  3.8640372  by  1.2479603,  retaining 
only  seven  decimal  places  in  the  product. 

EXPLANATION. — It  is  evident  that  we  need  OPERATION. 

regard    only    that    portion    of    each    partial  «  «  1  i  Q  7  9 

product  which  affects  the  figures  in  and  above  * 
the  seventh  decimal  place. 


Beginning  with  the   highest  figure  of  the  38640372 

multiplier,  we  obtain  the  first  partial  product.  7728074 

Taking  the  second  figure  of  the  multiplier,  we  1545614 

carry  each  figure  of  the  partial  product  one  270482 

place  to  the  right,  so  that  figures  of  the  same  34776 
order   shall  be  in   the   same   column.      This 
product  is  carried  out  one  place  further  than 


is   required,  so  as   to  secure  accuracy  in  the          4.8221650 

seventh  place,  and   we  draw  a  perpendicular 

line  to  separate  this  portion.     The  product  of  .04  by  the  right-hand 


110  KAY'S  HIGHER  ARITHMETIC. 

figure  in  the  seventh  place,  would  extend  to  the  ninth  place  of 
decimals ;  so  we  may  reject  the  last  figure,  and  commence  with  the  7. 
With  each  succeeding  figure  of  the  multiplier,  we  commence  to 
multiply  at  that  figure  of  the  multiplicand  which  will  produce  a 
product  in  the  eighth  place.  It  is  also  convenient  to  place  each 
figure  of  the  multiplier  directly  over  the  first  figure  of  the  multipli- 
cand taken.  In  multiplying  by  .007,  we  have  7  X  3  —  21 ;  but,  if  we 
had  been  expressing  the  complete  work,  we  should  have  5  to  carry 
to  this  place ;  the  corrected  product  is  therefore  21  +  5  =  26.  The 
product  from  the  last  figure,  3,  is  carried  two  places  to  the  right.  In 
the  total  product,  the  eighth  decimal  is  dropped ;  but  the  seventh 
decimal  figure  is  corrected  by  the  amount  carried. 

Rule. — 1.  Multiply  only  such  figures  as  shall  produce  one 
more  than  the  required  number  of  decimal  places. 

2.  Begin  with  the  highest  order  of  the  multiplier;   under  the 
right-hand  figure  of  each  partial  product,  place  the  right-hand 
figure  of  the   succeeding   one.      In   obtaining  such   right-hand 
figure,  let  that  number  be  added  which  would  be  carried  from 
multiplying  the  figure  of  the  next  lower  order. 

3.  Add  the  partial  products,  and  reject  the  right-hand  figure. 

KEMARKS. — 1.  It  will  be  found  convenient  to  write  the  multiplier 
in  a  reverse  order,  with  its  units1  figure  under  that  decimal  figure  of 
the  multiplicand  whose  order  is  next  lower  than  the  lowest  required. 
Thus,  in  the  fourth  example,  the  8  would  be  written  under  the  1. 

2.  In  carrying  the  tens  for  what  is  left  out  on  the  right,  carry  also 
one  ten  for  each  5  of  units  in  the  omitted  part ;  thus,  1  ten  for  5  or 
14  units,  3  tens  for  25  or  26  units,  etc.  Make  the  same  correction 
for  the  final  figure  rejected  in  the  product. 


EXAMPLES  FOR  PRACTICE. 

1.  Multiply  27.653  by  9.157,  preserving    three  decimal 
places.  253.219 

2.  Multiply  43.2071  by  3.14159,  preserving  four  decimal 
places.  135.7390 

3.  Multiply  3.62741   by  1.6432,  preserving  four  decimal 
places.  5.9606 


DIVISION  OF  DECIMALS.  Ill 

4.  9.012X48.75,  preserving  one  place.  439.3 

5.  4.804136  X  .010759,  preserving  six  places.      .051688 

6.  814T5TV  X  26f f ,  preserving  three  places.        21813.475 

7.  702.61  X  1.258-J&,  preserving  three  places.       884.020 

8.  849.93|  X  .0424444,  preserving  three  places.     36.075 

9.  880.695  X  131.72  true  to  units.  116005 

10.  .025381  X  .004907,  preserving  five  places.  .00012 

11.  64.01082  X  .03537,  preserving  six  places.       2.264063 

12.  1380.37^  X  -234f,  preserving  two  places.  324.16 


DIVISION  OF  DECIMALS. 

164.  Division  of  Decimals  is  the  process  of  finding  the 
quotient  when  either  or  when  each  term  is  a  decimal. 

165.  Since   the  dividend   corresponds  to  the  product  in 
multiplication  (Art.  73),  and  the  decimal  places  in  the  divi- 
dend are  as  many  as  in  both   factors  (Art.    162,  Prin.), 
we  derive  the  following  principles : 

PRINCIPLES. — 1.  The  dividend  must  contain  as  many  deci- 
mal places  as  the  divisor ;  and  when  both  have  the  same  number, 
the  quotient  is  an  integer. 

2.  The  quotient  must  contain  as  many  decimal  places  as 
the  number  of  those  in  the  dividend  exceeds  the  number  of  those 
in  the  divisor. 

PROBLEM.— Divide  .50312  by  .19 

OPERATION. 

.19). 50312(2. 64 8,  Am. 
SOLUTION. — The  division  is  per-  38 

formed  as  in  integers.     The  quo-  123 

tient     is     pointed     according     to  114 

Principle  2.      The  quotient  must  gY 

have  5  —  2  =  3  places.  75 

152 
152 


112  RAY'S  HIGHER  ARITHMETIC. 

PROOF. — By  expressing  the  decimals  as  common  fractions 
we  have : 

iWoVV  ±  rVo  rf  iWoVo  X  W  =W*  =  2.648 
PROBLEM.— Divide  .36  by  .008 

SOLUTION. — The  dividend  has  a  less  num-  OPERATION. 

her   of    decimal   places    than    the    divisor.  .008 )  .  3  60 
Annex    one    cipher,    making    the    number  45,  Ans. 

equal.     The  quotient  is  an  integer. 

PROBLEM.— Divide  .002Jf  by  .06f 

OPERATION-. 

SOLUTION.— Keducing  the  .002  J$  =  . 002475 

mixed   decimals    to   equiva-  .  0  6  f    =.066 

-lent  pure  decimals,  we  have  .  0  6  6  ) .  00  2  4  7  5  (.037  5,  Ans. 

.002475  and  .066.     Dividing,  198 

we    find   one   more    decimal  495 

place  necessary  to  make  the  462 

division  exact ;  and,  pointing  330 

by  Principle  2,  we  have  .0375  330 

Rule. — Divide  as  in  whole  numbers,  and  point  off  as  many 
decimal  places  in  the  quotient  as  those  in  the  dividend  exceed 
those  in  the  divisor. 

NOTE. — When  the  division  is  not  exact,  annex  ciphers  to  the 
dividend,  and  carry  the  work  as  far  as  may  be  necessary. 


EXAMPLES  FOR  PRACTICE. 


1.  63-^-4000.  .01575 

2.  3.15-^375.  .0084 

3.  1.008  -T-  18.  .056 

4.  4096-^.0.32  128000. 

5.  9.7-^-97000.  .0001 

6.  .  9 -f-.  00075  1200. 


7.  13-^78.12^  .1664 

8.  12.9-^-8.256         1.5625 

9.  81. 2096 -~  1.28     63.445 

10.  1-MOO.  .01 

11.  10.1 -M7.         .59412- 

12.  . 001 -f- 100.  .00001 


DIVISION  OF  DECIMALS.  113 

13.  12755  -f-  81632.  .15625 

14.  2401 -f- 21. 4375  112. 

15.  21.13212  -J-.916  23.07 

16.  36.72672^.5025  73.088 

17.  2483.25  -f-  5. 15625  481.6 

18.  142.0281 -T- 9.2376  15.375 

19.  .OSi-^.121  -66f  =  f. 

20.  .0001 -f-. 01  .0°! 

21.  95.3 -f-. 264  360.984848  + 

22.  1000 +  .001  1000000. 

23.  Ten  +- 1  tenth.  100. 

24.  .000001 +  .01  .0001 

25.  .00001  +  1000.  .00000001 

26.  16.275 +  .41664  39.0625 

27.  1  ten-millionth  ~  1  hundredth.  .00001 

166.  Oughtred's  Method. — If  the  quotient  is  not  re- 
quired to  contain  figures  below  a  certain  denomination,  the 
work  may  sometimes  be  abridged. 

PROBLEM.— Divide  84.27  by  1.27395807,  securing  a  quo- 
tient true  to  four  places  of  decimals. 

OPERATION. 

SOLUTION.— Since  1.2  jfr  jJ  ^  $ )  8  4.2  7  0  6  0  (  6  6.1  4  8  3  — 

the  divisor  is  greater  7643736 

than  1   and  less  than  2,  the  quotient  783264 

will  contain  six  places, — two   of  in-  764374 

tegers    and    four   of    decimals.     The  .18890 

highest  denomination  of  the  divisor,  12740 

multiplied  by  the  lowest   denornina-  6150 

tion  of  the  quotient,  would  obtain  a  5096 

figure  in  the  fourth  place.     We  take  1054 

one  place  more  as  in  multiplication  1019 

(Art.  163),  and  also  cut  off  two  figures  ~~r~ 

of  the  divisor,  since  these  can  not  affect  ~  ^ 
the  quotient  above  the  fourth  place. 

After  obtaining  the  first  figure  of  the  quotient,  we  drop  one  right- 
hand  figure  of  the  divisor  for  each  figure  obtained.      To  prevent 

errors,  we  cancel  the  figure  before  each  division. 
H.  A.  10. 


114  RAY'S  HIGHER  ARITHMETIC. 

Rule. — Find  the  figure  of  the  dividend  that  would  result  from 
multiplying  a  unit  in  the  highest  denomination  of  the  divisor  by  a 
unit  of  the  lowest  denomination  required  in  the  quotient.  Take 
one  more  figure  of  the  dividend  to  secure  accuracy.  Cut  off  any 
figures  of  the  divisor  not  needed  for  the  abbreviated  dividend. 

Divide  as  usual  until  the  figures  remaining  in  the  dividend 
are  all  divided.  At  each  subsequent  division,  drop  a  figure 
from  the  divisor,  carrying  the  number  necessary  from  the  product 
of  the  figure  omitted. 

Continue  until  the  divisor  is  reduced  to  two  figures. 

REMARK. — In  the  quotient,  5  units  of  an  omitted  order  may  be 
taken  as  1  unit  of  the  next  higher  order. 


EXAMPLES  FOR  PRACTICE. 

1.  1000 -f-. 98,  preserving  two  places.  1020.41 

2.  6215.75  -r-  .99^,  preserving  three  places.  6246.985 

3.  28012 -f- .993,  preserving  two  places.  28209.47 

4.  52546. 35  -^.99f,  preserving  three  places.  52678.045 

5.  4840  -f-  .9875,  preserving  two  places.  4901.27 

6.  2  -^  1.4142136,  preserving  seven  places.  1.4142135 

7.  9.869604401  -f- 3. 14159265,  preserving  eight  places. 

3.14159265 


Topical   Outline. 
DECIMAL  FRACTIONS. 


Definitions. 
Decimal  Point. 

C  Pure. 
Classes J  Mixed. 

(_  Complex. 
Principles. 
Numeration,  Rule. 
Notation,  Rule. 


Reduction.  /  L  Dec- to  a  Com- 

I II.  Com.  Fraction  to  a  Dec. 

Addition,  Rule. 
Subtraction,  Rule. 
Multiplication,  Rule. 

Abbreviated  Multiplication. 
Division,  Rule. 

Abbreviated   J>i vision. 


X.  CIRCULATING  DECIMALS. 

167.  In  reducing  common  fractions  to  decimals,  the 
process,  in  some  cases,  does  not  terminate.  This  gives  rise 
to  Circulating  Decimals. 

PRINCIPLE  I.  —  If  any  prime  factors  oilier  than  2  and  5  are 
found  in  the  denominator  of  a  fraction  in  its  lowest  terms,  the 
resulting  decimal  will  be  interminate. 

DEMONSTRATION.  —  If  the  fraction  is  in  its  lowest  terms,  the 
numerator  and  denominator  are  prime  to  each  other  (131,  Hem.  2). 
In  the  process  of  reduction,  the  numerator  is  multiplied  by  10.  By 
this  means  the  factors  2  and  5  may  be  introduced  into  the  numera- 
tor as  many  times  as  necessary  ;  but  no  others  are  introduced. 
Therefore,  if  any  factors  other  than  2  and  5  are  found  in  the  denom- 
inator, the  division  can  not  be  made  complete,  and  the  resulting 
decimal  will  be  interminate. 

Thus,  ^  =  2X2X2X32X2X2X5  -  .009375 


but>    6^0  =  23^1^5  =-116666+ 

It  is  evident  that  the  first  will  terminate  if  the  numerator  be  mul- 
tiplied six  times  by  10,  carrying  the  decimal  to  the  sixth  place.  In 
the  same  way  we  reduce  ffc  to  a  decimal  containing  four  places. 
But  since  the  factor  3  is  found  in  the  denominator  of  ^,  the  frac- 
tion can  not  be  exactly  reduced,  though  the  numerator  be  multi- 
plied by  any  power  of  10. 

« 

PRINCIPLE  II.  —  Every  interminate  decimal  arising  from  the 
reduction  of  a  common  fraction  will,  if  the  division  be  carried 
far  enough,  contain  the  same  figure,  or  set  of  figures,  repeated 
in  the  same  order. 

(115) 


116  RAY'S  HIGHER  ARITHMETIC. 

DEMONSTRATION. — Each  of  the  remainders  must  be  less  than  the 
denominator  which  is  used  as  the  divisor  (Art.  78,  Note  2).  If  the 
division  be  carried  far  enough,  some  remainder  must  be  found  equal 
to  some  remainder  already  found,  and  the  subsequent  figures  in  the 
quotient  must  be  similar  to  the  figures  found  from  the  former 
remainder. 

Thus,  in  reducing  \,  we  find  the  decimal  .142857,  and  then  have 
the  remainder  1,  the  number  we  started  with;  if  we  annex  a  cipher, 
we  shall  get  1  for  the  next  figure  of  the  quotient,  4  for  the  next,  etc. 

168.  1.  Interminate  decimals,  on  this  account,  have 
received  the  name  of  Circulating  or  Recurring  Decimals. 

2.  A  Circulate    or    Circulating    Decimal    has    one    or 

more  figures  constantly  repeated  in  the  same  order. 

3.  A  Repetend  is  the  figure  or  set  of  figures  repeated, 
and  it  is  expressed  by  placing  a  dot  over  the  first  and  last 
figure;    thus,   -f  =  .  142857;  if  there  be  one  figure  repeated, 
the  dot  is  placed  over  it,  thus,  -f  —  .6666  +  =  «6 

4.  A  Pure  Circulate  has  no  figures  but  the  repetend ;  as, 
.5  and  .124 

5.  A    Mixed    Circulate    has    other    figures    before    the 
repetend;  as,  .2083  and  .31247 

6.  A  Simple  Repetend  has  one  figure;  as,  .4 

7.  A  Compound  Repetend  has  two  or  more  figures ;  as, 
.59 

8.  A  Perfect  Repetend  is  one  which  contains  as  many 
decimal  places  as  there  are  units  in  the  denominator,  less  1 ; 
thus,  |  =  . 142857 

9.  Similar  Repetends  begin  and  end  at  the  same  deci- 
mal place;  as,  .427  and  .536 

10.  Dissimilar  Repetends  begin  or  end  at  different  deci- 
mal places;  as,  .205  and  .312468 


CIRCULATING  DECIMALS.  117 

11.  Conterminous  Repetends  end  at  the    same   place; 
as,  .50397  and  .42618 

12.  Co-originous  Repetends   begin  at  the    same  place ; 
as,  .5  and  .124 

169.  Any  terminate  decimal  may  be  considered  a   circu- 
late, its  repetend   being  ciphers;  as,  .35  =  .350  =  .350000 
Any  simple   repetend   may   be   made    compound,    and   any 
compound  repetend  still  more  compound,  by  taking  in  one 
or  more  of  the  succeeding  repetends;  as,  .3  =  .33333,  and 
.0562  =  .056262,  and  .257  =  .257257257 

REMARKS. — 1.  When  a  repetend  is  thus  enlarged,  be  careful  to 
take  in  no  part  of  a  repetend  without  taking  the  whole  of  it ;  thus,  if 
we  take  in  2  figures  iif  the  last  example,  the  result,  .25725,  would  be 
incorrect,  for  the  next  figure  understood  being  7,  shows  that  25725 
is  not  repeated. 

2.  A  repetend  may  be  made  to  begin  at  any  lower  place  by  carry- 
ing its  dots  forward,  each  the  same  distance ;  thus,  .5  =  .555,  and 
.2941  ==  .29414,  and  5.1836  —  5.183683 

3.  Dissimilar  repetends  can  be  made  similar,  by  carrying  the  dots 
forward  till  they  all  begin  at  the  same  place  as  the  one  farthest 
from  the  decimal  point. 

4.  Similar  repetends  may  be  made  conterminous  by  enlarging  the 
repetends  until  they  all  contain  the  same  number  of  figures.     This 
number  will  be  the  least  common  multiple  of  the  numbers  of  figures 
in  the  given  repetends. 

For,  suppose  one  of  the  repetends  to  have  2,  another  3,  another 
4,  and  the  last  6  figures ;  in  enlarging  the  first,  figures  must  be  taken 
in,  2  at  a  time,  and  in  the  others,  3,  4,  and  6  at  a  time. 

170.  Circulating  decimals  originate,  as  has  been  already 
shown,    in    changing   some  common    fractions   to    decimals. 
Then,  having  given  a  circulate,  it  can  always  be  changed  to 
an  equivalent  common  fraction. 

171.  Circulating   decimals    may    be    added,    subtracted, 
multiplied,  and  divided  as  other  fractions. 


118  RAY'S  HIGHER  ARITHMETIC. 


CASE   I. 

172.    To    reduce    a    pure    circulate    to   a   common 
fraction. 

PROBLEM. — Change  .53  to  a  common  fraction. 

OPERATION. 

SOLUTION. — Kemoving  the  100  times  the  repetend  —  5  3.5  3 
decimal  point  one  place  to  the  Once  the  repetend  =  .53 

right,  multiplies  the  repetend      /.  99  times  the  repetend  =  53. 
by  10 ;  two  places,  by  100,  and  Once  the  repetend  =  f  f ,  Ans. 

so  on.     Then,  multiplying  by 

100,  and  subtracting  the  repetend  from  the  product,  removes  all  of 
that  part  to  the  right  of  the  decimal  point,  and,  dividing  53  by  99, 
we  have  the  common  fraction  which  produced  the  given  repetend. 

« 

PROBLEM. — Change  .456  to  a  common  fraction. 

OPERATION. 

1000  times  the  repetend  =  4  5  6.4  5  6 
Once  the  repetend  —          .456 
999    times  the  repetend  =  456. 
.*.  once  the  repetend  =  f  f  f ,  .4ns. 

PROBLEM. — Change  25.6  to  a  common  fraction. 

OPERATION. 

Carry  the  dot  forward  thus  :  2  5.6  =  2  5.6  2  5 

1000  times  the  repetend  =  6  2  5.6  2  5 
Once  the  repetend  =         .625 
999    times  the  repetend  =  625. 
.'.  once  the  repetend  =  f  f  f 
Whence  the  25.625  =  2  5  fff,  Am. 

NOTE. — From  these  solutions  the  following  rule  is  derived. 

Rule. —  Write  the  repetend  for  the  numerator,  omitting  the 
decimal  point  and  the  dots,  and  for  the  denominator  write  as 
many  ffs  as  there  are  figures  in  the  repetend,  and  reduce  the 
fraction  to  its  lowest  terms. 


CIRCULATING  DECIMALS.  119 

CASE    II. 

173.    To   reduce  a   mixed   circulate   to  a   common 
fraction. 

PROBLEM. — Change  .821437  to  a  common  fraction. 

OPERATION. 

821  --7 
Omitting  the  decimal  point,  we  have :       .821437  = p*  = 

821X999  +  437  _  82  1  (  1  OOP  — 1 )  +  437 
1000X999  999000 

8  2  1000  —  8  21  +  437   820616   102577 


999000        999000   124875 
Or,  briefly,     8  2143  7  —  821   102577 


,  Ans. 


999000     124875' 


Ans. 


PROBLEM. — Change  .048  to  a  common  fraction. 

OPERATION. 

Omitting  the  decimal  point,  we  have : 

048  =  -!!-       4  8     _4(10-1)         8 

100       100^900  900        r900 

_36_       _8 .M___li 

900  +  900~900~225'     m' 
Or,  briefly,        48  —  4       44        11 
900     ~900~225' 

NOTE. — The  following  rule  is  derived  from  the  preceding  solu- 
tions. 

«* . 

Rule.  1.  For  tlie  numerator,  subtract  the  part  which  precedes 
the  repetend  from  the  whole  expression,  both  quantities  being  con- 
sidered as  units. 

2.  For  the  denominator,  write  as  many  9's  as  there  are  figures 
in  the  repetend,  and  annex  as  many  ciphers  as  there  are  decimal 
figures  before  each  repetend. 


120 


BAY'S  HIGHER   ARITHMETIC. 


Keduce  to  common  fractions : 


1. 
2. 
3. 
4. 
5. 
6. 


.3 

.05 

.123 

2.63 

.31 

.0216 

48.1 


i- 


TV 


rfs'- 


9. 
10. 
11. 
12. 
13. 
14. 


1.001 

.138 

.2083 

85.7142 

.063492 

.4476190 

.09027 


1  9  0  9  ' 

A- 


85f 

eV 
T4oV 


ADDITION  OF  CIRCULATES. 

174.  Addition  of  Circulates  is  the  process  of  finding 
the  sum  of  two  or  more  circulates.  Similar  circulates  only 
can  be  added. 

PROBLEM.— Add  .256,  5.3472,  24.815,  and  .9098 


SOLUTION. — Make  the  circulates  similar. 
The  first  column  of  figures  which  would 
appear,  if  the  circulates  were  continued,  is 
the  same  as  the  first  figures  of  the  repe- 
tends,  6,  7,  1,  0,  whose  sum,  14,  gives  1  to  be 
carried  to  the  right-hand  column.  Since 
the  last  six  figures  in  each  number  make  a 
figures  of  the  sum  also  make  a  repetend. 


OPERATION. 

.2566666666 

5.3  4  7  2  7  2  7  2  7  2 

2  4.8  158158158 

.9098000000 

31.3295552097 
repetend,  the  last  six 


Rule.— Make  the  repetends  similar,  if  they  be  not  so;  add, 
and  point  off  as  in  ordinary  decimals,  increasing  the  right-hand 
column  by  the  amount,  if  any,  which  would  be  carried  to  it  if 
the  circulates  were  continued;  then  make  a  repetend  in  the  sum, 
similar  to  those  above. 


REMARK. — In  finding  the  amount  to  be  carried  to  the  right-hand 
column,  it  may  be  necessary,  sometimes,  to  use  the  two  succeeding 
figures  in  each  repetend. 


SUBTRACTION  OF  CIRCULATES.  121 


EXAMPLES  FOR  PRACTICE. 

1.  Add  .453,  .068,  .327,  .946  1.796 

2.  Add  3.04,  6.456,  23.38,   .248  33.1334 

3.  Add  .25,  .104,  .61,  and  .5635  1.536 

4.  Add  1.03,  .257,  5.04,  28.0445245  34.37 

5.  Add  .6,  .138,  .05,  .0972,  .0416  1. 

6.  Add  9.21107,  .65,  5.004,  3.5622  18.43 

7.  Add  .2045,  .09,  and  .25  .54 

8.  Add  5.0770,  .24,  and  7.124943  12.4 

9.  Add  3.4884,  1.637,  130.81,  .066  136.00 


SUBTRACTION  OF  CIECULATES. 

175.  Subtraction  of  Circulates  is  the  process  of  find- 
ing the  difference  between  two  circulates.  The  two  circu- 
lates must  be  similar. 

PROBLEM.— Subtract  9.3i56  from  12.9021 

OPERATION. 

SOLUTION. — Prepare  the  numbers  for  sub-  1  2.9  0212121 
traction.  If  the  circulates  were  continued,  the  9.3  1  5  6  1  5  6  1 

next  figure   in  the   subtrahend    (5)    would   be  358650559 

larger    than   the   one  above  it  (2)  ;    therefore, 
carry  1  to  the  right-hand  figure  of  the  subtrahend. 

Rule. — Make  the  repetends  similar,  if  they  be  not  so ;  subtract 
and  point  off  as  in  ordinary  decimals,  carrying  1,  however, 
to  the  right-hand  figure  of  the  subtrahend,  if  on  continuing  the 
circulates  it  be  found  necessary ;  then  make  a  repetend  in  the 
remainder,  similar  to  those  above. 

REMARK. — It  may  be  necessary  to  observe  more  than  one  of  the 
succeeding  figures  in  the  circulates,  to  ascertain  whether  1  is  to  be 

carried  to  the  right-hand  figure  of  the  subtrahend  or  not. 
H.  A.  11. 


122  HAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  Subtract  .0074  from  .26  .259 

2.  Subtract  9.09  from  15.35465  6.25 

3.  Subtract  4.51  from  18.23673  13.72 

4.  Subtract  37.0128  from  100.73  63.71 

5.  Subtract  8.27  from  10.0563  1.7836290 

6.  Subtract  190.476  from  199.6428571  9.16 

7.  Subtract  13.637  from  104.1  90.503776 


MULTIPLICATION  OF  CIKCULATES. 

176.  Multiplication  of  Circulates  is  the  process  of 
finding  the  product  when  either  or  when  each  of  the  factors 
is  a  circulate. 

PROBLEM.— Multiply  .3754  by  17.43 

SOLUTION. — In    forming    the    partial  OPERATION. 

products,  carry  to  the  right-hand  figures  .3754 
of  each  respectively,  the  numbers  1,  3,  0,         1  7.4  3  —  1  7.4  £ 

arising  from   the    multiplication   of   the  1501777 

figures  that  do  not  appear.     The  repetend  2628lili 

of  the  multiplier  being  equal  to  J,  ^  of  37544444 

the  multiplicand  is  125148,  whose  figures  125148 

are  set  down  under  those  of  the  multipli- 

,    ,  i>  i.    it:  ^  •     j  6.5452481  Ans. 

cand   from   which    they   were    obtained. 

Point  the  several  products,  carry  them  forward  until  their  repetends 
are  similar,  and  add  for  answer. 

Rule. — 1.  If  the  multiplier  contain  a  repetend,  change  it  to  a 
common  fraction. 

2.  Then  multiply  as  in  multiplication  of  decimals,  and  add 
to   the    right-lxind   figure    of    each  partial  product  the  amount 
necessary  if  the  repetend  were  repeated. 

3.  Make  the  partial  products  similar,  and  find  their  sum. 


DIVISION  OF  CIRCULATES.  123 


EXAMPLES  FOR  PRACTICE. 

1.  4.735  X  7.349  34.800H3 

2.  .07067  X  .9432  .066665 

3.  714.32X3.456  2469.173814 

4.  16.204  X  32.75  530.810446 

5.  19.072  X  .2083  3.97348 

6.  3.7543  X  4.7157  17.7045082 

7.  1.256784  X  6.42081  8.069583206 


DIVISION  OF  CIKCULATES. 

177.  Division  of  Circulates  is  the  process  of  finding 
the  quotient  when  either  or  when  each  of  the  terms  is  a 
circulate. 

PROBLEM.  —  Divide  .154  by  .2 
OPERATION. 


Rule.  —  Change  tJie  terms  to  common  fractions;  then  divide 
as  in  division  of  fractions,  and  reduce  the  result  to  a  repetend. 

REMARK.  —  This  is  the  easiest  method  of  solving  problems  in  di- 
vision of  circulates.  The  terms  may  be  made  similar,  however,  and 
the  division  performed  without  changing  the  circulates  to  common 
fractions. 

EXAMPLES  FOR  PRACTICE. 

1.  .75  +  .1  6.81 

2.  51.491  -M  7.  3.028 

3.  681.5598879-;-  94.  7.2506371 

4.  90.5203749-^6.754  13.401 

5.  11.  068735402-^.245  45.13 

6.  9.5330663997  -f-  6.217  1.53 

7.  3.500691358024-^7.684  .45 


124 


RAY'S  HIGHER  ARITHMETIC. 


Topical  Outline. 
CIRCULATING  DECIMALS. 


1.  Principles. 


2.  Definitions.... 


1.  Circulate. 

2.  Repetend. 

3.  Pure  Circulate. 

4.  Mixed  Circulate. 

5.  Simple  Repetend. 

6.  Compound  Repetend. 

7.  Perfect  Repetend. 

8.  Similar  Repetends. 

9.  Dissimilar  Repetends. 

10.  Conterminous  Repetends. 

11.  Co-originous  Repetends. 


3.  Reduction f  Case  I. 

lease  II. 


f  Definition. 

4.  Addition  

-j    Rule. 
V  Applications. 

C  Definition. 

5.  Subtraction  

-j    Rule. 

I  Applications. 

f  Definition. 

6.  Multiplication 

-j    Rule. 

I  Applications. 

c  Definition. 

7.  Division  

-j    Rule. 

(.  Applications. 

XL  COMPOUND  DENOMINATE 
NUMBERS. 

178.  1.  A  Measure  is  a  standard  unit  used  in  estimating 
quantity.     Standard  units  are  fixed  by  law  or  custom. 

2.  A  quantity  is  measured  by  finding  how  many  times  it 
contains  the  unit. 

3.  Denomination  is  the  name  of  a  unit  of  measure  of  a 
concrete  number. 

4.  A    Denominate    Number    is    a     concrete    number 
which  expresses  a  particular  kind  of  quantity;  as,  3  feet,  7 
pounds. 

5.  A  Compound  Denominate  Number  is  one  expression 
of  a  quantity  by  different   denominations  under   one  kind  of 
measure;  as,  5  yards,  2  feet,  and  8  inches. 

6.  All  measures  of  denominate  numbers  may  be  embraced 
under    the    following    divisions :     Value,    Weight,  Extension, 
and  Time. 

MEASUEES  OF  VALUE. 

179.  1.  Value  is  the  worth  of  one  thing  as  compared 
with  another. 

2.  Value  is  of  three   kinds:    Intrinsic,    Commercial,  and 
Nominal. 

3.  The  Intrinsic  Value  of  any  thing  is  measured  by  the 
amount  of  labor  and  skill  required  to  make  it  useful. 

4.  The  Commercial  Value  of  any  thing  is  its  purchasing 
power,  exchangeability,  or  its  worth  in  market. 

(125) 


126  RAY'S  HIGHER  ARITHMETIC. 

5.  The  Nominal  Value  is  the  name  value  of  any  thing. 

6.  Value  is  estimated  among  civilized  people  by  its  price 
in  money. 

7.  Money  is  a  standard  of  value,  and  is  the  medium  of 
exchange;    it  is  usually  stamped  metal,    called   coins,  and 
printed  bills  or  notes,  called  paper  money. 

8.  The  money  of  a  country  is  its  Currency.     Currency 
is  national  or  foreign. 

United  States  Money. 

180.  United  States  Money  is  the  legal  currency  of  the 
United  States.  It  is  based  upon  the  decimal  system;  that 
is,  ten  units  of  a  lower  order  make  one  of  the  next  higher. 

The  Dollar  is  the  unit.  The  same  unit  is  the  standard 
of  Canada,  the  Sandwich  Islands,  and  Liberia. 

TABLE. 

10  mills,  marked  m.,  make  1  cent,  marked  ct. 

10  cents  "      1  dime,       "        d. 

10  dimes  "      1  dollar,     "        $. 

10  dollars  "      1  eagle,      "        E. 


NOTE.  —  The  cent  and  mitt,  which  are  TJo  and  Y^nr  °f  a  dollar, 
derive  their  names  from  the  Latin  centum  and  mille,  meaning  a  hun- 
dred and  a  thousand  ;  the  dime,  which  is  -^  of  a  dollar,  is  from  the 
French  word  disme,  meaning  ten. 

REMARKS.  —  1.  United  States  money  Was  established,  by  act  of 
Congress,  in  1786.  The  first  money  coined,  by  the  authority  of  the 
United  States,  was  in  1793.  The  coins  first  made  were  copper  cents. 
In  1794  silver  dollars  were  made.  Gold  eagles  were  made  in  1795; 
gold  dollars,  in  1849.  Gold  and  silver  are  now  both  legally  standard. 
The  trade  dollar  was  minted  for  Asiatic  commerce 

2.  The  coins  of  the  United  States  are  classed  a&  bronze,  nickel,  silver, 


MEASURES  OF  VALUE. 


127 


and  gold.     The  name,  value,  composition,  and  weight  of  each  coin 
are  shown  in  the  following  table : 


TABLE. 


COIN. 

VALUE. 

COMPOSITION. 

WEIGHT. 

BRONZE. 

One  cent. 

1  cent. 

95  parts  copper,  5  parts  tin  &  zinc. 

48      grains  Troy. 

NICKEL. 

3-cent  piece. 

3  cents. 

75  parts  copper,  25  parts  nickel. 

30     grains  Troy. 

5-cent  piece. 

5  cents. 

75     "           "       25     " 

77.16      " 

SILVER. 

Dime. 

10  cents. 

90  parts  silver,  10  parts  copper. 

2.5  grams. 

Quarter  dollar. 

25  cents. 

90     "         "       10      " 

6.25      " 

Half  doilar. 

50  cents. 

90     "         "       10      " 

12.5 

Dollar. 

100  cents. 

90     "         "       10      " 

412.5  grains  Troy. 

GOLD. 

Dollar. 

100  cents. 

90  parts  gold,  10  parts  copper. 

25.8  grains  Troy. 

Quarter  eagle. 

iyz  dollars. 

90     "        "      10     " 

64.5 

Three  dollar. 

3  dollars. 

90     "        "       10     " 

77.4 

Half  eagle. 

5  dollars. 

90     "        "       10     " 

129 

Eagle. 

10  dollars. 

90     "        "      10     " 

258 

Double  eagle. 

20  dollars. 

90     "        "       10      " 

516 

3.  A  deviation  in  weight  of  J  a  grain  to  each  piece,  is  allowed  by- 
law in  the  coinage  of  Double  Eagles  and  Eagles ;  of  £  of  a  grain  in 
the  other  gold  pieces;  of  1|  grains  in  all  silver  pieces  ;  of  3  grains 
in  the  five-cent  piece ;  and  of  2  grains  in  the  smaller  pieces. 

4.  The  mill  is  not  coined.     It  is  used  only  in  calculations. 

I 
181.     In    reading  U.  S.  Money,  name  the  dollars  and  all 

higher  denominations  together  as  dollars,  the  dimes  and  cents  as 
cents,  and  the  next  figure,  if  there  be  one,  as  mills ; 

Or,  name  the   whole   number  as   dollars,  and   the  rest   as  a 
decimal  of  a  dollar. 

Thus,  $9.124  is  read  9  dollars  12  ct.  4  mills,  or  9  dollars  124 
thousandths  of  a  dollar. 


128  RAY'S  HIGHER  ARITHMETIC. 


English  or  Sterling  Money. 

182.  English   or  Sterling   Money  is  the   currency  of 
the  British  Empire. 

The  pound  sterling  (worth  $4.8665  in  TL  8.  money)  is 
the  unit,  aud  is  represented  by  the  sovereign  and  the  £i 
bank-note. 

TABLE. 

4  farthings,  marked  qr.,  make  1  penny,  marked  d. 
12  pence  "     1  shilling,      "        s. 

20  shillings  "     1  pound,  ^    "       £. 

EQUIVALENT  TABLE. 

<£          s.  d.  qr. 

1  =  20  =  240  =  960. 

1=    12=    48. 

1  =      4. 

REMARKS. — 1.  The  abbreviations,  <£,  s.,  d.,  q.,  are  the  initials 
of  the  Latin  words  libra,  solidarius,  denarius,  quadrans,  signifying, 
respectively,  pound,  shilling,  penny,  and  quarter. 

2.  The  coins  are  gold,  silver,  and  copper.     The  gold  coins  are  the 
sovereign  (  =  <£!),  half  sovereign   ( =  10s.),  guinea  (  —  21s.),  and 
half-guinea  (  =  10s.  6d.)      The  silver  coins   are  the  crown  (  =  5s.), 
the  half  crown  (  =  2s.  6d.),  the  florin  (  —  2s.),  the  shilling,  and  the 
six-penny,  four-penny,  and  three- penny  pieces.     The  penny,  half- 
penny, and  farthing  are  the  copper  coins.      The  guinea,  half-guinea, 
crown,  and  half-crown  are  no  longer  coined,  but  some  of  them  are 
in  circulation. 

3.  The  standard  for  gold  coins  is  ^J  pure  gold  and  -^  alloy ;  the 
standard  for  silver  is  f  J  pure    silver  and  •£-$  copper ;    pence  and 
half-pence  are  pure  copper. 

French  Money. 

183.  French    Money  is   the   legal   currency  of  France. 
Its  denominations  are  decimal. 


MEASURES  OF   VALUE.  129 

The  franc  (worth  19.3  cents  in  U.  S.  money)  is  the  unit. 
The  franc  is  also  used  in  Switzerland  and  Belgium,  and. 
under  other  names,  in  Italy,  Spain,  and  Greece. 

TABLE. 

10  millimes,  marked  m.,  make  1  centime,  marked  c. 
10  centimes  "      1  decime,  d. 

10  decimes  "      1  franc,  "      fr. 

EQUIVALENT  TABLE. 

1  fr.  =  10  d.  —  100  c.  =  1000  m. 

1  d.  =    10  c.  =    100  m. 

1  c.  =      10  m. 

REMARK. — The  coins  are  of  gold,  silver,  and  bronze.  The  gold 
coins  are  100,  50,  20,  10,  and  5-franc  pieces ;  they  are  .9  pure  gold. 
The  20-franc  piece  weighs  99.55  gr.  The  silver  coins  are  5,  2,  and 
1 -franc  pieces,  and  50  and  20-centime  pieces  ;  they  are  now  .835 
pure  silver.  The  bronze  coins  are  10,  5,  2,  and  1 -centime  pieces. 
The  decime  is  not  used  in  practice.  All  sums  are  given  in  francs 
and  centimes,  or  hundredths. 


German  Money. 

184.  German  Money  is  the  legal  currency  of  the  Ger- 
man Empire. 

The  mark,  or  reichsmark  (worth  23.8  cents  in  U.  S. 
money),  is  the  unit.  The  only  other  denomination  is  the 
pfennig  (penny). 

TABLE. 

100  pfennige,  marked  Pf.,  make  1  mark,  marked  EM. 

EEMARK.— The  coins  are  of  gold,  silver,  and  copper.  The  gold 
coins  are  of  the  value  of  40,  20,  10,  and  5  marks;  the  silver  coins,  3, 
2,  and  1-mark  pieces,  and  50,  20,  and  10  pfennige.  The  copper,  2  and 
1 -pfennig  pieces.  Gold  and  silver  are  .9  fine. 


130  RA  Y'S  HIGHER  ARITHMETIC. 


MEASURES  OF  WEIGHT. 

185.  Weight  is  the  measure  of  the  force  called  gravity, 
which  draws  bodies  toward  the  center  of  the  earth. 

The  standard  unit  of  weight  in  the  United  States  is  the 
Troy  pound  of  the  Mint. 

186.  Three   kinds  of  weight  are  in  use, — Troy    Weight, 
Apothecaries'   Weight,  and  Avoirdupois    Weight. 

Troy  Weight. 

187.  Troy  Weight  is   used    in    weighing   gold,    silver, 
platinum,  and  jewels.     Formerly  it  was  used  in  philosoph- 
ical and  chemical  works. 

TABLE. 

24  grains,  marked  gr.,  make  1  pennyweight,  marked  pwt. 
20  pwt.  "      1  ounce,  "        oz. 

12  oz.  "      1  pound,  Ib. 

EQUIVALENT  TABLE. 

ft).         oz.         pwt.  gr. 

1  =  12  =  240  =  5760. 
1  =    20  =    480. 
1  =      24. 

EEMARK. — The  Troy  pound  is  equal  to  the  weight  of  22.7944 
cubic  inches  of  pure  water  at  its  maximum  density,  the  barometer 
being  at  30  inches.  The  standard  pound  weight  is  identical  with  the 
Troy  pound  of  Great  Britain. 

Apothecaries'  Weight. 

188.  Apothecaries'  Weight  is  used  by  physicians  and 
apothecaries    in    prescribing    and    mixing    dry    medicines. 
Medicines  are  bought  and  sold  by  Avoirdupois  Weight. 


MEASURES  OF  WEIGHT.  131 

TABLE. 

20  grains,  marked  gr.,  make  1  scruple,  marked  9. 

39  "  1  dram,             "       3. 

83  "  1  ounce,            "       g. 

12  §  "  1  pound,           "      ft. 

EQUIVALENT  TABLE. 

n>.        3-        3-         9-  gr. 

1  =  12  =  96  =  288  —  5760. 
1  =    8  =    24  =    480. 
1-5=      3  =      60. 

1   r=        20. 

REMARK. — The  pound,  ounce,  and  grain  of  this  weight  are  the 
same  as  those  of  Troy  weight ;  the  pound  in  each  contains  12  oz.  = 
5760  gr. 

Avoirdupois  or  Commercial  Weight. 

189.  Avoirdupois  or  Commercial  Weight  is  used  for 
weighing  all  ordinary  articles. 

TABLE. 

16  ounces,  marked  oz.,  make  1  pound,  marked  Ib. 

25  Ib.  "      i  quarter,  ".  qr. 

4  qr.  "      1  hundred-weight,  "  cwt. 

20  cwt.  "      1  ton,  "  T. 

EQUIVALENT  TABLE. 

T.        cwt.          qr.  Ib.  oz. 

I  =  20  =^  80  =  2000  =  32000. 

1  =  4  =  100  =  1600. 

1  =   25  =   400. 

1  =    16. 


132  RA  Y'S  HIGHER  ARITHMETIC. 

KEMARKS. — 1.  In  Great  Britain,  the  qr.  —  28  lb.,  the  cwt.  =  112 
lb.,  the  ton  =  2240  lb.  These  values  are  used  at  the  United  States 
custom-houses  in  invoices  of  English  goods,  and  are  still  used  in 
some  lines  of  trade,  such  as  coal  and  iron. 

2.  Among  other  weights  sometimes   mentioned  in  books,  are  :  1 
stone,  horseman's  weight,  =  14  lb.;  1  stone  of  butcher's  meat  =  8  lb.; 
1  clove  of  wool  =  7  lb. 

3.  The  lb.  avoirdupois  is  equal  to  the  weight  of  27.7274  cu.  in.  of 
distilled  water  at  62°(Fahr.);  or  27.7015  cu.  in.  at  its  maximum 
density,  the  barometer  at  30  inches.     For  ordinary  purposes,  1  cubic 
foot  of  water  can  be  taken  62|  lb.  avoirdupois. 

4.  The  terms  gross  and  net  are  used  in  this  weight.     Gross  weight 
is  the  weight  of  the  goods,  together  with  the  box,  cask,  or  whatever 
contains  them.     Net  iveight  is  the  weight  of  the  goods  alone. 

5.  The  word  avoirdupois  is  from  the  French  avoirs,  du,  pois,  signify- 
ing goods  of  weight. 

6.  The  ounce  is  often  divided  into  halves  and  quarters  in  weigh- 
ing.    The  sixteenth  of  an  ounce  is  called  a  dram. 


COMPARISON   OF   WEIGHTS. 

190.  The  pound  Avoirdupois  weighs  7,000  grains  Troy, 
and  the  Troy  pound  weighs  5,760  grains,  hence  there  are 
1,240  grains  more  in  the  Avoirdupois  pound  than  in  the 
Troy  pound. 

The  following  table  exhibits  the  relation  between  certain 
denominations  of  Avoirdupois,  Troy,  and  Apothecaries' 
Weight. 

Avoirdupois.       Troy.        Apothecaries'. 
1  lb.   —  1TV¥  lb.      =  IfV*  ». 
1  oz.  =     lif  oz.     =     Hf  g. 

1  lb.    =       1  ib. 
1  oz.     =         1  §. 
1  gr.     —         1  gr. 
1  pwt.  =        I  z- 

1  pwt.  =  \\  3. 


MEASURES  OF  EXTENSION.  133 

KEMAKK. — In  addition  to  the  foregoing,  the  following,  called 
Diamond  Weight,  is  used  in  weighing  diamonds  and  other  precious 
stones. 

TABLE. 

16  parts  make  1  carat  grain  =.792  Troy  grains. 

'4  carat  grains      "      1  carat  =3.168    " 

NOTE. — This  carat  is  entirely  different  from  the  assay  carat,  which 
has  reference  to  the  fineness  of  gold.  The  mass  of  gold  is  considered 
as  divided  into  twenty-four  parts,  called  carats,  and  is  said  to  be  so 
many  carats  fine,  according  to  the  number  of  twenty-fourths  of  pure 
gold  which  it  contains. 


MEASUKES  OF  EXTENSION. 

191.  1.  Extension  is  that  property  of  matter  by  which 
it  occupies  space.     It  may  have  one  or  more  of  the  three 
dimensions, — length,  breadth,  and  thickness. 

2.  A  line  has  only  one  dimension, — length. 

3.  A  surface  has  two  dimensions,  length  and  breadth. 

4.  A  solid    or   volume    has   three   dimensions, — length, 
breadth,  and  thickness. 

192.  Measures  of  Extension  embrace  : 

f  Long  Measure. 

1.  Linear  Measure.  J  Chain  Measure. 

Mariners'  Measure. 
I  Cloth  Measure. 

2.  Superficial  Measure.  /  ^uare  Mea*ure' 

(  burveyors   Measure. 

3.  Solid  Measure.  ( Liquid  Measure. 

4.  Measures  of  Capacity.  <  Apothecaries1  Measure. 

5.  Angular  Measure.          (Dry  Measure. 


134  HA  Y'S  HIGHER  ARITHMETIC. 


Long  or  Linear  Measure. 

193.  Linear  Measure  is  used  in  measuring  distances,  or 
length,  in  any  direction. 

The  standard  unit  for  all  measures  of  extension  is 
the  yard,  which  is  identical  with  the  Imperial  yard  of 
Great  Britain. 

TABLE. 

12     inches,  marked  in.,  make  1  foot,  marked  ft. 

3     ft.  "  1  yard,       "  yd. 

5^  yd.  or  16J  ft.  "  1  rod,         "  rd. 

320     rd.  "  1  mile,        "  mi. 

EQUIVALENT  TABLE. 

mi.         rd.             yd.                 ft.  in. 

1  =  320  =  1760  =  5280  =  63360. 

1  =   5i  =   161  =  198. 

l-.-.:=    3  =  36. 

1  =  12. 

REMARKS. — 1.  The  standard  yard  of  the  United  States  was 
obtained  from  England  in  1856.  It  is  of  bronze,  and  of  due  length 
at  59.8°  Fahr.  A  copy  of  the  former  standard  is  deposited  at 
each  state  capital :  this  was  about  y^V o  °f  an  incn  too  long. 

2.  The  rod  is  sometimes  called  perch  or  pole.     The  furlong,  equal  to 
40  rods,  is  seldom  used. 

3.  The  inch  may  be  divided  into  halves,  fourths,  eighths,  etc.,  or 
into  tenths,  hundredths,  etc. 

4.  The  following  measures   are  sometimes  used : 

• 

12  lines  make  1  inch. 

3  barleycorns        "  1     " 

3  inches  "  1  palm. 

4  inches                 "  1  hand. 
9  inches                 "  1  span. 

18  inches  "      1  cubit. 

3  feet  "      1  pace. 


MEASURES  OF  EXTENSION.  135 

194.  Chain  Measure  is  used  by  surveyors  in  measuring 
land,  laying  out  roads,  establishing  boundaries,  etc. 

TABLE. 

7.92  inches,  marked  in.,  make  1  link,  marked  li. 
100  li.  "      1  chain,     "        ch. 

80  ch.  "      1  mile,       "         mi. 

EQUIVALENT  TABLE. 

mi.        ch.  li.  in. 

1  =  80  =  8000  =  63360. 

1  =    100  =      792. 

1  =     7.92. 

REMARKS. — 1.  The  surveyors'  chain,  or  Gunter's  chain,  is  4  rods, 
or  66  feet  in  length.  Since  it  consists  of  100  links,  the  chains  and 
links  may  he  written  as  integers  and  hundredths ;  thus,  2  chains  56 
links  are  written  2.56  ch. 

2.  The  engineers'  chain  is  100  feet  long,  and  consists  of  100  links. 

3.  The  engineers'  leveling  rod  is  used  for  measuring  vertical  dis- 
tances.     It  is  divided  into  feet,  tenths,   and   hundredths,  and,   by 
means  of  a  vernier,  may  be  read  to  thousandths. 

195.  Mariners'  Measure  is  used  in  measuring  the  depth 
of  the  sea,  and  also  distances  on  its  surface. 

TABLE. 

6  feet  make  1  fathom. 
720  feet      "      1  cable-length. 

REMARKS. — 1.  A  nautical  mile  is  one  minute  of  longitude,  meas- 
ured on  the  equator  at  the  level  of  the  sea.  It  is  equal  to  1.152J 
statute  miles.  60  nautical  miles  =  1  degree  on  the  equator,  or  69.16 
statute  miles.  A  league  is  equal  to  3  nautical  miles,  or  3.458  statute 
miles. 

2.  Depths  at  sea  are  measured  in  fathoms ;  distances  are  usually 
measured  in  nautical  miles. 


136  KAY'S  HIGHER  ARITHMETIC. 

196.  Cloth   Measure  is  used   in    measuring  dry-goods. 
The  standard  yard  is  the  same  as  in  Linear  Measure,  but 
is  divided   into  halves,    quarters,   eighths,    sixteenths,    etc.,    in 
place  of  feet  and  inches. 

REMARKS. — 1.  There  was  formerly  a  recognized   table  for  Cloth 
Measure,  but  it  is  now  obsolete.     The  denominations  were  as  follows : 
2}  inches,  marked  in.,  make  1  nail,  marked  na. 
4     na.  or  9  in.  "      1  quarter,  "       qr. 

4    qr.  "      1  yard,       "       yd. 

2.  At  the  custom-house,  the  yard  is  divided  decimally. 

Superficial   or  Surface  Measure. 

197.  1.  Superficial  Measure  is  used  in  estimating  the 
numerical  value  of  surfaces;  such  as,  land,  weather-board- 
ing, plastering,  paving,  etc. 

2.  A  surface  has  length  and  breadth,  but  not  thickness. 

3.  The  area  of  a  surface  is  its  numerical  value;  or  the 
number  of  times  it  contains  the  measuring  unit. 

4.  A  superficial  unit  is  an  assumed  unit  of  measure  *for 
surfaces. 

Usually  the  square,  whose  side  is  the  linear  unit,  is  the 
unit  of  measure ;  as,  the  square  inch,  square  foot,  square 
yard. 

5.  A  Rectangle   may  be  defined   as 
a  surface  bounded  by  four  straight  lines 
forming  four  square  corners;  as  either  of 
the  figures  A,  B. 

6.  When  the  four  sides  are  equal  the 
rectangle    is    called    a    square;    as    the 
figure  C. 

7.  The  area  of  a  rectangle  is  equal  to  its  length  multiplied  by 
its  breadth. 


MEASURES  OF  EXTENSION.  137 

EXPLANATION. — Take  a  rectangle  4  inches  long 
by  3  inches  wide.  If  upon  each  of  the  inches  in 
the  length,  a  square  inch  be  conceived  to  stand, 
there  will  be  a  row  of  4  square  inches,  extending 
the  whole  length  of  the  rectangle,  and  reaching  1 
inch  of  its  width.  At*  the  rectangle  contains  as  many  such  rows  as 
there  are  inches  in  its  width,  its  area  must  be  equal  to  the  number 
of  square  inches  in  a  row  (4)  multiplied  by  the  number  of  rows  (3), 
=  12  square  inches.  This  statement  (7),  as  commonly  understood, 
can  present  no  exception  to  Prin.  2,  Art.  60. 

TABLE. 

144  square  inches  (sq.  in.)  make  1  square  foot,  marked  sq.  ft. 
9  sq.  ft.  "     1  square  yard,       "      sq.  yd. 

30^  sq.  yd.  "     1  square  rod,         "      sq.  rd. 

160  sq.  rd.  "     1  acre,  "      A. 

EQUIVALENT  TABLE. 

A.       sq.  rd.       sq.  yd.  sq.  ft.  sq.  in. 

1  =  160  =  4840  =  43560  =  6272640. 

1  =   30J  =  •  2721  =   39204. 

1  =    9  =   '  1296. 

1  =    144. 

NOTE. — The  following,  though  now  seldom  used,  are  often  found 
in  records  of  calculations  : 

40  perches  (P.),  or  sq.  rds.,  make  1  rood,  marked  B. 
4  roods  "      1  acre,         "        A. 

198.     Surveyors'    Measure    is    a    kind    of    superficial 
measure,  which  is  used  chiefly  in  government  surveys. 

TABLE. 

625  square  links  (sq.  li.)  make  1  square  rod,  sq.  rd. 

16  sq.  rd.  "      1  square  chain,  sq.  ch. 

10  sq.  ch.  "      1  acre,  A. 

640  A.  u      1  square  mile,  sq.  mi. 

36  sq.  mi.  (6  miles  square)  "      1  township,       J?~p«~~ 
H.  A.  12. 

/T         ^    OF  THE 
/     IIKIIWCTDQITY 


138  RAY'S  HIGHER  ARITHMETIC. 


EQUIVALENT  TABLE. 

Tp.    sq.  mi.  A.  sq.  ch.  sq.  rd.  sq.  li. 

1  =  36  =-23040  =  230400  =  3686400  =  2304000000. 

1  =   640  =   6400  =  102400  =   64000000. 

1  =    10  =    160  =    100000. 

1  =     16  =     10000. 

1  =       625. 

Solid   Measure. 

199.     A  Solid  has  length,  breadth,  and  thickness. 

Solid  Measure  is  used  in  estimating  the  contents  or  volume 
of  solids. 

A  Cube  is  a  solid,  bounded  by  six  equal  squares,  called 
faces.  Its  length,  breadth,  and  thickness  are  all  equal. 

REMARK. — The  size  or  name  of  any  cube,  like  that  of  a  square, 
depends  upon  its  side,  as  cubic  inch,  cubic  foot,  cubic  yard. 

EXPLANATION. — If  each  side  of  a  cube 
is  1  inch  long,  it  is  called  a  cubic  inch ;  if 
each  side  is  3  feet  (1  yard)  long,  as  repre- 
sented in  the  figure,  it  is  a  cubic  or  solid 
yard. 

When  the  base  of  a  cube  is  1  square  yard, 
it  contains  3X3=9  square  feet ;  and  1  foot 
high  on  this  base,  contains  9  solid  feet ;  2 
feet  high  contains  9  X  2  =  18  solid  feet  ;  3  feet  high  contains  9X3 
=:  27  solid  feet.  Also  it  may  be  shown  that  1  solid  or  cubic  foot 
contains  12  X  12  X  12  —  1728  solid  or  cubic  inches. 

The  unit  by  which  all  solids  are  measured  is  a  cube, 
whose  side  is  a  linear  inch,  foot,  etc.,  and  their  size  or 
solidity  will  be  the  number  of  times  they  contain  this  unit. 

REMARK. — The  simplest  solid  is  the  rectangular  solid,  which  is 
bounded  by  six  rectangles,  called  its  faces,  each  opposite  pair  being 
equal,  and  perpendicular  to  the  other  four;  as,  for  example,  the 


MEASURES  OF  EXTENSION.  139 

ordinary  form  of  a  brick  or  a  box  of  soap.  If  the  length,  breadth, 
and  thickness  are-  the  same,  the  faces  are  squares,  and  the  solid  is 
a  cube. 

TABLE. 

1728  cubic  inches  (cu.  in.)  make  1  cubic  foot,  cu.  ft. 
27  cu.  ft.  "1  cubic  yard,  cu.  yd. 

EQUIVALENT  TABLE. 

cu.  yd.     cu.  ft.       cu.  in. 

1  =  27  =  46656. 
1  ==     1728. 

REMARKS. — 1.  A  perch  of  stone  is  a  mass  16|  ft.  long,  1^  ft.  wide, 
and  1  ft.  high,  and  contains  24J  cu.  ft. 

2.  Earth,  rock-excavations,  and  embankments    are   estimated  by 
the  cubic  yard. 

3.  Round  timber  will  lose  \  in  being  sawed,  hence  50  cubic  ft.  of 
round   timber  is    said  to  be    equal  to   40  cubic  ft.  of  hewn  timber, 
which  is  a  ton. 

4.  Fire-wood  is  usually  measured  by  the  cord.     A  pile  of  wood  4 
ft.  high,  4  ft.  wide,  and  8  ft.  long,  contains  128  cubic  feet  or  one  cord. 
One  foot  in  length  of  this  pile,  or  16  cu.  ft.,  is  called  a  cord  foot. 

5.  Planks   and  scantling  are  estimated  by  board  measure.     In  this 
measure,  1  reduced  foot,  1  ft.  long,  1  ft.  wide,  and  1  in.  thick,  contains 
12  X  12  X  1  =144  cu.  in.     All  planks  and  scantling  less  than  an 
inch  thick,  are  reckoned   at  that  thickness  ;  but,  if  more  than  an 
inch  thick,  allowance  must  be  made  for  the  excess. 


Measures  of  Capacity. 

200.  Capacity  means  room  for  things. 

Measures    of  Capacity  are    divided    into    Measures   of 
Liquids  and  Measures  of  Dry  Substances. 

201.  Liquid  Measure  is  used  in  measuring  liquids,  and 
in  estimating  the  capacities  of  cisterns,  reservoirs,  etc. 

The  gallon,  which    contains  231  cu.   in.,  is  the  unit  of 
measure  in  liquids. 


140  RA  YJ  S  HIGHER  ARITHMETIC. 

NOTE.  —  This  gallon  of  231  cubic  inches  was  the  standard  in 
England  at  the  time  of  Queen  Anne.  The  present  imperial  gallon 
of  England  contains  10  Ib.  of  water  at  62°  Fahr.,  or  277.274  cubic 
inches. 

TABLE. 

4  gills,  marked  gi.,    make  1  pint,       marked  pt. 
2  pt.  "      1  quart,  "        qt. 

4  qt.  "      1  gallon,         "        gal. 

EQUIVALENT  TABLE. 

gal.       qt.        pt.         gi. 

1  =  4  =  8  =  32. 

1  =  2  =    8. 


NOTE.  —  Sometimes  the  barrel  is  estimated  at  31  J  gal.,  and  the 
hogshead  at  63  gal.;  but  usually  each  package  of  this  description 
is  gauged  separately. 

202.  Apothecaries'  Fluid  Measure  is  used  for  meas- 
uring all  liquids  that  enter  into  the  composition  of  medical 
prescriptions. 

TABLE. 

60  minims,  marked  TV,  make  1  fluid  drachm,  marked  f%. 

8  f  3                                       "      1  fluid  ounce,  "        f§. 

16  f£                                       *      1  pint,  "         O. 

8  O                                       "      1  gallon,  "    cong. 

EQUIVALENT  TABLE. 

cong.     O.         f§.  £3.  nt. 

1  =  8  =  128  =  1024  =  61440. 

1  =     16  =     128  =    7680. 

1  =        8  =      480. 


MEASURES  OF  EXTENSION.  141 

NOTES.  —  1.  Cong,  is  an  abbreviation  for  congiarium,  the  Latin  for 
gallon  ;  O.  is  the  initial  of  octans,  the  Latin  for  one  eighth,  the  pint 
being  one  eighth  of  a  gallon. 

2.  For  ordinary  purposes,  1  tea-cup  =  2  wine-glasses  =  8  table- 
spoons =  32  tea-spoons  =  4  f  §. 

203.  Dry  Measure  is  used  for  measuring  grain,  fruit, 
vegetables,  coal,  salt,  etc. 

The  Winchester  bushel  is  the  unit  ;  it  was  formerly  used 
in  England,  and  so  called  from  the  town  where  the  standard 
was  kept.  It  is  8  in.  deep,  and  18%  in.  in  diameter, 
and  contains  2150.42  cu.  in.,  or  77.6274  Ib.  av.  of  distilled 
water  at  maximum  density,  the  barometer  at  30  inches. 

NOTE.  —  This  bushel  was  discarded  by  Great  Britain  in  1826,  and 
the  imperial  bushel  substituted;  the  latter  contains  2218.192  cu.  in., 
or  eighty  pounds  avoirdupois  of  distilled  water. 

TABLE. 

2  pints,  marked  pt.,  make  1  quart,  marked  qt. 
8  qt.  "      1  peck,         "        pk. 

4  pk.  "      1  bushel,      "        bu. 

EQUIVALENT  TABLE. 

bu.       pk.         qt.  pt. 

1  =  4  =  32  =  64. 
1  =    8  =  16. 


EEMARKS.  —  1.  4  qt.  or  J  peck  =  1  dry  gal.  =  268  8  cu.  in.  nearly. 

2.  The  quarter  is  still  used  in  England  for  measuring  wheat,  of 
which  it  holds  eight  bushels,  or  480  pounds  avoirdupois. 

3.  When  articles  usually  measured  by  the  above  table  are  sold 
by  weight,  the  bushel  is  taken  as  the  unit.     The  following  table  gives 
the   legal   weight  of  a  bushel  of   various   articles  in   avoirdupois 
pounds  : 


142  RAY'S  HIGHER  ARITHMETIC. 

TABLE. 


ARTICLES. 

LB. 

EXCEPTIONS. 

Beans. 

60 

Me.,  64;  N.  Y.,  62. 

Coal. 

80 

f  Ohio,  70  of  cannel  ;  Ind.,  70  mined  out  of 
\     the  state  ;  Ky.,  76  of  anthracite. 

Corn  (Indian). 

56 

N.  Y.,  58  ;  CaL,  52  ;  Arizona,  54. 

Flax  Seed. 

56 

N.  Y.  andN.  J.,  55;  Kan.,  54. 

Oats. 

32 

|  Md.,  26  ;  Me.,  N.  H.,  N.  J.,  Pa.,  30  ;  Neb.,  34  ; 
\     Montana,  35  ;  Oregon  and  Wash.,  36. 

Potatoes  (Irish). 

60 

Ohio,  58  ;  Wash.,  50. 

Eye. 

56 

La.,  32  ;  Cal.,  54. 

IMass.,  70  ;  Pa.,  coarse,  85  ;  ground,  70  ;  fine. 

Salt. 

50 

62;    Ky.  and  111.,  fine,  55;    Mich.,  56; 

Col.  and  Dak.,  80. 

Wheat. 

60 

COMPARATIVE  TABLE  OF  MEASURES. 


cu.  in.  gal.     cu.  in.  qt. 
Liquid  Measure,  231  57f 

Dry  Measure  (J  pk.),     2684  67^ 


cu.in.pt.     cu.  in.gi. 

28£  7^ 

33|  8f 


Angular  or  Circular  Measure. 

204.     A  plane  angle  is  the  difference  of  direction  of  two 
straight  lines  which  meet  at  a  point. 

EXPLANATION. — Thus,  the  two  lines  AB  and  AC 
meet  at  the  point  A,  called  the  apex.  The  lines  AB 
and  AC  are  the  sides  of  the  angle,  and  the  difference 
in  direction,  or  the  opening  of  the  lines,  is  the  angle 
itself. 

Angular  Measure  is  used  to  measure  angles,  directions, 
latitude,  and  longitude,  in  navigation,  astronomy,  etc. 

A  circle   is  a  plane  surface  bounded  by  a  line,  all  the 
points  of  which  are  equally  distant  from  a  point  within. 


MEASURE  OF  TIME.  143 

EXPLANATIONS. — The  bounding  lino 
ADBEA  is  a  circumfertnce.  Every  point 
of  this  line  is  at  the  same  distance  from 
the  point  C,  which  is  called  the  center. 
The  circle  is  the  area  included  within  the 
circumference.  Any  straight  line  drawn 
from  the  center  to  the  circumference  is 
called  a  radius;  thus,  CD  and  CB  are 
radii.  Any  part  of  the  circumference,  as 
AEB  or  AD,  is  an  arc.  A  straight  line, 

like  AB,  drawn  through  the  center,  and  having  its  ends  in  the  cir- 
cumference, is  a  diameter;  it  divides  the  circle  into  two  equal  parts. 

NOTES. — 1.  Every  circumference  contains  360  degrees ;  and,  the 
apex  of  an  angle  being  taken  as  the  center  of  a  circle,  the  angle  is 
measured  by  the  number  of  degrees  in  the  arc  included  by  the  sides 
of  the  angle. 

2.  The  angle  formed  by  two  lines  perpendicular  to  each  other,  as 
the  radii  AC  and  DC  in  the  above  figure,  is  a  right  angle,  and  is 
measured  by  the  fourth  part  of  a  circumference,  90°,  called  a  quadrant. 

TABLE. 

60  seconds,  marked  ",  make  1  minute,  marked,    '. 

60'  "      1  degree,  "         °. 

360°  "      1  circumference,      "         c. 

EQUIVALENT  TABLE. 
c.          o  /  // 

1  =  360  =  =  21600  =  1296000. 
1  =        60  =        3600. 
1  =  60. 

NOTE. — The  twelfth  part  of  a  circumference,  or  30°,  is  called  a 
sign. 

»     MEASURE  OF  TIME. 

205.     1.  Time  is  a  measured  portion  of  duration. 

2.  A  Year  is  the  time   of  the   revolution    of  the   earth 


144  RA  Y>S  HIGHER  ARITHMETIC. 

around  the  sun  ;  a  Day  is  the  time  of  the  revolution  of  the 
earth  on  its  axis. 

3.  The  Solar  Day  is  the  interval  of  time  between  two 
successive  passages  of  the  sun  over  the  same  meridian. 

4.  The  Mean  Solar  Day  is  the  mean,  or  average,  length 
of  all  the  solar  days  in  the  year.     Its  duration  is  24  hours, 
and  it  is  the  unit  of  Time  Measure. 

5.  The  Civil  Day,  used  for  ordinary  purposes,  commences 
at  midnight  and  closes  at  the  next  midnight. 

6.  The  Astronomical  Day  commences  at  noon  and  closes 
at  the  next  noon. 

TABLE. 

60  seconds,  marked  sec.,  make  1  minute,  marked  min. 

60  min.  "  1  hour,             "  hr. 

24  hr.  "  1  day,               "  da, 

7  da.  "  1  week,            "  wk. 

4  wk.  "  1  month,          "  mori. 

12  calendar  mon.  "  1  year,             "  yr. 

365  da.  "  1  common  year. 

366  da.  "  1  leap  year. 

100  yr.  "      1  century,  marked  cen. 

NOTE.—  1  Solar  year  =  365  da.  5  hr.  48  min.  46.05  sec.  =  365£  da., 
nearly. 

EQUIVALENT  TABLE. 

yr.       mo.         wk.  da.  hr.  min.  sec. 

365  =  876°  ==  525^00  ==  31536000. 


.   12  =  52  = 

"  366  =  8784  —  527040  ==  31622400. 

1  =    7  =  168  —  10080  =   604800. 

1  =   24  =   1440  =   86400. 

1  =    60  =    3600. 

1  =      60. 


MEASURE  OF  TIME.  145 

NOTE. — The  ancients  were  unable  to  find  accurately  the  number 
of  days  in  a  year.  They  had  10,  afterward  12,  calendar  months, 
corresponding  to  the  revolutions  of  the  moon  around  the  earth.  In 
the  time  of  Julius  Caesar  the  year  contained  365|-  days ;  instead  of 
taking  account  of  the  \  of  a  day  every  year,  the  common  or  civil 
year  was  reckoned  365  days,  and  every  4th  year  a  day  was  inserted 
(called  the  intercalary  day),  making  the  year  then  have  366  days. 
The  extra  day  was  introduced  by  repeating  the  24th  of  February, 
which,  with  the  Romans,  was  called  the  sixth  day  before  the  kalends  of 
March.  The  years  containing  this  day  twice,  were  on  this  account 
called  bissextile,  which  means  having  two  sixths.  By  us  they  are  gen- 
erally called  leap  years. 

But  365J  days  (365  days  and  6  hours)  are  a  little  longer  than  the 
true  year,  which  is  365  days  5  hours  48  minutes  46.05  seconds.  The 
difference,  11  minutes  13.95  seconds,  though  small,  produced,  in  a 
long  course  of  years,  a  sensible  error,  which  was  corrected  by 
Gregory  XIII.,  who,  in  1582,  suppressed  the  10  days  that  had  been 
gained,  by  decreeing  that  the  5th  of  October  should  be  the  15th. 

206.  To  prevent  difficulty  in  future,  it  has  been  decided 
to  adopt  the  following  rule. 

Rule  for  Leap  Years. — Every  year  that  is  divisible  by  4 
is  a  leap  year,  unless  it  ends  with  two  ciphers;  in  which  case 
it  must  be  divisible  by  400  to  be  a  leap  year. 

ILLUSTRATION.— Thus,  1832,  1648,  1600,  and  2000  are  leap  years ; 
but  1857,  1700,  1800,  1918,  are  not. 

NOTES. — 1.  The  Gregorian  calendar  was  adopted  in  England  in  1752. 
The  error  then  being  11  days,  Parliament  declared  the  3d  of  September 
to  be  the  14th,  and  at  the  same  time  made  the  year  begin  January 
1st,  instead  of  March  25th.  Eussia,  and  all  other  countries  of  the 
Greek  Church,  still  use  the  Julian  calendar;  consequently  their 
dates  (Old  /Style)  are  now  12  days  later  than  ours  (New  Style).  The 
error  in  the  Gregorian  calendar  is  small,  amounting  to  a  day  in 
3600  years. 

2.  The  year  formerly  began  with  March  instead  of  January ;  con- 
sequently, September,  October,  November,  and  December  were  the  7th, 
8th,  9th,  and  10th  months,  as  their  names  indicate;  being  derived 

from  the  Latin  numerals  Septeui  (7),  Octo  (8),  Novem  (9),  Decem  (10). 
H.  A.  13. 


146  RAY1 8  HIGHER  ARITHMETIC. 


COMPARISON  OF  TIME  AND   LONGITUDE. 

207.  The  longitude  of  a  place  is  its  distance  in  degrees, 
minutes,  and  seconds,  east  or  west  of  an  established  meridian. 

NOTE. — The  difference  of  longitude  of  two  places  on  the  same  side  of 
the  established  meridian,  is  found  by  subtracting  the  less  longitude 
from  the  greater;  but,  of  two  places  on  opposite  sides  of  the  meridian, 
the  difference  of  longitude  is  found  by  adding  the  longitude  of  one  to 
the  longitude  of  the  other. 

The  circumference  of  the  earth,  like  other  circles,  is 
divided  into  360  equal  parts,  called  degrees  of  longitude. 

The  sun  appears  to  pass  entirely  round  the  earth,  360°, 
once  in  24  hours,  one  day;  and  in  1  hour  it  passes  over 
15°.  (360° +- 24  =  15°.) 

As  15°  equal  900',  and  1  hour  equals  60  minutes  of 
time,  therefore,  the  sun  in  1  minute  of  time  passes  over  15' 
of  a  degree.  (900'  -j-  60  =  15'.) 

As  15'  equal  900",  and  1  minute  of  time  equals  60  seconds 
of  time,  therefore,  in  1  second  of  time  the  sum  passes  over 
15"  of  a  degree.  (900"  -f-  60  =  15".) 

TABLE  FOR  COMPARING  LONGITUDE  AND  TIME. 

15°  of  longitude  =  1  hour  of  time. 
15'  of  longitude  =  1  min.  of  time. 
15"  of  longitude  =  1  sec.  of  time. 

NOTE, — If  one  place  has  greater  east  or  less  west  longitude  than 
another,  its  time  must  be  later ;  and,  conversel}7,  if  one  place  has 
later  time  than  another,  it  must  have  greater  east  or  less  west 
longitude. 

MISCELLANEOUS  TABLES. 

208.  The  words  folio,  quarto,  octavo,  etc.,  used  in  speak- 
ing of  books,  show  how  many  leaves  a  sheet  of  paper  makes. 


THE  METRIC  SYSTEM. 


147 


A  sheet  folded  into 
2  leaves,  called  a     folio, 


12 
16 
32 


a     quarto  or  4to, 
an  octavo  or  8vo, 
a     duodecimo  or  12rno, 
a     16mo, 
a     32mo, 


makes     4  pages. 
«         3       <c 

16  " 

"  24  " 

"  32  " 

64  " 


Also, 


24  sheets  of  paper  make  1  quire. 
20  quires  "      1  ream. 

2  reams  "      1  bundle. 

5  bundles  "      1  bale. 


12  things  make  1  dozen. 

12  dozens  or  144  things     "      1  gross. 

12  gross  or  144  dozens 

20  things 

56  Ib. 
100  Ib. 
196  Ib. 
200  Ib. 


1  great  gross. 
1  score. 

1  firkin  of  butter. 
1  quintal  of  fish. 
1  bbl.  of  flour. 
1  bbl.  of  pork. 


THE  METKIC  SYSTEM. 
Historical. 

The  Metrio  System  is  an  outgrowth  of  the  French  Eevolution  of 
1789.  At  that  time  there  was  a  general  disposition  to  break  away 
from  old  customs  ;  and  the  revolutionists  contended  that  every  thing 
needed  remodeling.  A  commission  was  appointed  to  determine 
an  invariable  standard  for  all  measures  of  length,  area;  solidity, 
capacity,  and  weight.  After  due  deliberation,  an  accurate  survey 
was  made  of  that  portion  of  the  terrestrial  meridian  through 
Paris,  between  Dunkirk,  France,  and  Barcelona,  Spain ;  and  from 
this,  the  distance  on  that  meridian  from  the  equator  to  the  pole 
was  computed.  The  quadrant  thus  obtained  was  divided  into  ten 
million  equal  parts ;  one  part  was  called  a  meter,  and  is  the  base 
of  the  system.  From  it  all  measures  are  derived. 


•~o 


iO 


148  RAY'S  HIGHER  ARITHMETIC. 

In  1795  the  Metric  System  was  adopted  in  France.  It  is  now 
used  in  nearly  all  civilized  countries.  It  was  author- 
ized by  an  act  of  Congress  in  the  United  States  in  1866. 

209.  The  Metric  System  is  a  decimal  system 
of  weights  and  measures. 

The  meter  is  the  primary  unit  upon  which 
the  system  is  based,  and  is  also  the  unit  of 
length.  It  is  39.37043  inches  long,'  which  is 
very  nearly  one  ten-millionth  part  of  the  distance 
on  the  earth's  surface  from  the  equator  to  the 
§  pole,  as  measured  on  the  meridian  through  Paris. 

I 

REMARK. — The  standard  meter,  a  bar  of  platinum,  is 

kept  among  the  archives  in  Paris.     Duplicates  of  this 

bar  have  been  furnished  to  the  United  States. 

H 
fe 

O 

210.  The  names  of  the  lower  denominations 
in    each    measure    of   the    Metric    System    are 
formed   by  prefixing    the    Latin   numerals,    deci 
(.1),  centi  (.01),  and   milli  (.001)  to  the  unit  of 
that  measure ;  those  of  the  higher  denominations, 
by   prefixing    the    Greek    numerals,    deka   (10), 
hekto  (100),   kilo  (1000),  and  myria  (10000),  to 
the  same  unit. 

These  prefixes  may  be  grouped  about  the  unit 
of    measure,  showing    the  decimal  arrangement 
of  the  system,  as  follows: 

f  milli    =  .001 

Lower    Denominations.  «{  centi   —  .01 

I  deci     —  .1 

Unit  of  Measure  —          1. 
fdeka    =        10. 

Higher  Denotations.      J**  = 

I  myria  =  10000. 


THE  METRIC  SYSTEM.  149 

211.  The  units  of  the  various  measures,  to  which  these 
prefixes  are  attached,   are   as    follows : 

The  Meter,  which  is  the  unit  of  Length. 

The  Ar,  "      "     "      "      "  Surface. 

The  Liter,  "       (:     "      "      "  Capacity. 

The  Gram,  "       "     "      "      "  Weight. 

REMARK. — The  name  of  each  denomination  thus  derived,  imme- 
diately shows  its  relation  to  the  unit  of  measure.  Thus,  a  centi- 
meter is  one  one-hundredth  of  a  meter ;  a  kilogram  is  a  thousand 
grams ;  a  hektoliier  is  one  hundred  liters,  etc. 

* 

Measure  of  Length. 

212.  The  Meter  is  the  unit  of  Length,  and  is  the  de- 
nomination used  in  all  ordinary  measurements. 

TABLE. 


10  millimeters,  marked 

mm.,  make  1  centimeter,  marked  cm. 

10  centimeters 

"      1  decimeter, 

"       dm. 

10  decimeters 

"      1  meter, 

m. 

10  meters 

"      1  dekameter, 

"     Dm. 

10  dekameters 

'"      1  hektometer, 

"     Hm. 

10  hektometers 

"      1  kilometer, 

"     Km. 

10  kilometers 

"      1  myriameter, 

"     Mm. 

HEMARKS. — 1.  The  figure  on  page  148  shows  the  exact  length 
of  the  decimeter,  and  its  subdivisions  the  centimeter  and  millimeter. 

2.  The  centimeter  and  millimeter  are  most  often  used  in  meas- 
uring very  short  distances ;  and  the  kilometer,  in  measuring  roads 
and  long  distances. 

Measure  of  Surface. 

\ 

213.  The  Ar  (pro.  ar)  is  the  unit  of  Land  Measure ;  it 
is  a  square,  each  side  of  which  is  10  meters  (1  dekameter) 
in  length,  and  hence  its  area  is  one  square  dekameter. 


150  RAY'S  HIGHER  ARITHMETIC. 


TABLE. 

100  centars,  marked  ca.,  make  1  ar,     marked  a. 
100  ars  "      1  hektar,  "        Ha. 

REMARK. — The  square  meter  (marked  m2)  and  its  subdivisions  are 
used  for  measuring  small  surfaces. 


Measure  of  Capacity. 

214.  The  Liter  (pro.  le'ter)  is  the  unit  of  Capacity.  It 
is  equal  in  volume  to  a  cube  whose  edge  is  a  decimeter ; 
that  is,  one  tenth  of  a  meter. 

TABLE. 

10  milliliters,  marked  ml.,  make  1  centiliter,  marked  cl. 

10  centiliters  "  1  deciliter,          "        dl. 

10  deciliters  "  1  liter,               "        1. 

10  liters  "  1  dekaliter,        "        DL 

10  dekaliters  "  1  hektoliter,       "        HI. 

REMARKS. — 1.  This  measure  is  used  for  liquids  and  for  dry  sub- 
stances. The  denominations  most  used  are  the  liter  and  hektoliter ; 
the  former  in  measuring  milk,  wine,  etc.,  in  moderate  quantities,  and 
the  latter  in  measuring  grain,  fruit,  etc.,  in  large  quantities. 

2.  Instead  of  the  milliliter  and  the  kiloliter,  it  is  customary  to  use 
the  cubic  centimeter  and  the  cubic  meter  (marked  m3),  which  are 
their  equivalents. 

3.  For  measuring  wood  the  ster  (pro.  ster)  is  used.     It  is  a  cubic 
meter  in  volume. 


Measure  of  Weight. 

215.  The  Gram  (pro.  gram)  is  the  unit  of  Weight.  It 
was  determined  by  the  weight  of  a  cubic  centimeter  of  dis- 
tilled water,  at  the  temperature  of  melting  ice. 


THE  METRIC  SYSTEM.  151 


TABLE. 

10  milligrams,  marked  nig.,    make  1  centigram,  marked  eg. 

10  centigrams  "  1  decigram,  "      dg. 

10  decigrams  "  1  gram,  "        g. 

10  grams  "  1  dekagram,  "     Dg. 

10  dekagrams  "  1  hektogram,  "     Hg. 

10  hektograms  "  1  kilogram,  "     Kg. 

10  kilograms  "  1  myriagram,  "     Mg. 

10  myriagrams,  or  100  kilograms  "  1  quintal,  "       Q. 

10  quintals,  or  1000  "         "  1  metric  ton,  "    M.T. 

REMARKS. — 1.  The  gram,  kilogram  (pro.  kil'o-gram),  and  metric 
ton  are  the  weights  commonly  used. 

2.  The  gram  is  used  in  all  cases  where  great  exactness  is  required ; 
such   as,  mixing   medicines,  weighing  the  precious  metals,  jewels, 
letters,  etc. 

3.  The  kilogram,  or,  as  it  is  commonly  abbreviated,  the  "kilo," 
is  used  in  weighing  coarse  articles,  such  as  groceries,  etc. 

4.  The  metric   ton  is  used  in  weighing  hay  and  heavy  articles 
generally. 

216.  Since,  in  the  Metric  System,  10,  100,  1000,  etc., 
units  of  a  lower  denomination  make  a  unit  of  the  higher 
denomination,  the  following  principles  are  derived : 

PRINCIPLES. — 1.  A  number  is  reduced  to  a  LOWER  denom- 
ination by  removing  the  decimal  point  as  many  places  to  the 
RIGHT  as  there  are  ciphers  in  the  multiplier. 

2.  A  number  is  reduced  to  a  HIGHER  denomination  by 
removing  the  decimal  point  as  many  places  to  the  LEFT  as 
there  are  ciphers  in  the  divisor. 

ILLUSTRATIONS. — Thus,  15.03  m.  is  read  15  meters  and  3  centi- 
meters ;  or,  15  and  3  hundredths  meters.  Again,  15.03  meters  =  1.503 
dekameters  =  .1503  hektometer  =  150.3  decimeters  =  1503  centi- 
meters. As  will  be  seen,  the  reduction  is  effected  by  changing  the 
decimal  point  in  precisely  the  same  manner  as  in  United  States  Money. 


152 


RAY'S  HIGHER  ARITHMETIC. 


217.  The  following  table  presents  the  legal  and  approx- 
imate values  of  those  denominations  of  the  Metric  System 
which  are  in  common  use. 


TABLE. 


DENOMINATION. 

LEGAL   VALUE. 

APPROX.   VALUE. 

Meter. 

39.37  inches. 

3  ft.  3|  inches. 

Centimeter. 

.3937  inch. 

|  inch. 

Millimeter. 

.03937  inch. 

A-  inch. 

Kilometer. 

.62137  mile. 

f  mile. 

Ar. 

119.6  sq.  yards. 

4  sq.  rods. 

Hektar. 

2.471  acres. 

2J  acres. 

Square  Meter. 

1.196  sq.  yards. 

10|  sq.  feet. 

Liter. 

1.0567  quarts. 

1  quart. 

Hektoliter. 

2.8375  bushels. 

2  bu.  3J  pecks. 

Cubic  Centimeter. 

.061  cu.  inch. 

Y1^  cu.  inch. 

Cubic  Meter. 

1.308  cu.  yards. 

35J  cu.  feet. 

Ster. 

.2759  cord. 

\  cord. 

Gram. 

15.432  grains  troy. 

15|  grains. 

Kilogram. 

2.2046  pounds  av. 

2^  pounds. 

Metric  Ton. 

2204.6  pounds  av. 

1  T.  204  pounds. 

Topical   Outline. 
COMPOUND  NUMBERS. 


1.  Preliminary  Definitions. 


2.  Value 


1.  Definitions. 

2.  United  States  and  Canadian  Money. 

3.  English  Money. 

4.  French  Money. 

5.  German  Money. 


3.  Weight 


ri.  Troy  Weight. 

' -I  2.  Apothecaries'  Weight. 

l3.  Avoirdupois  Weight. 


TOPICAL  OUTLINE. 


153 


COMPOUND  NUMBERS. — ( Concluded.) 


f  1.  Linear... 


4.  Extension. 


2.  Superficial  Measure.... 

3.  Solid  Measure. 


ri.  Long  Measure. 
!  2.  Chain  Measure. 
)  3.  Mariners'  Measure. 
1 4.  Cloth  Measure. 

1.  Surface  Measure. 

2.  Surveyors'  Measure. 


4.  Measures  of  Capacity.  /  1-  Liquid. 
I  2. 


I  5.  Angular  Measure. 

5.  Time. 

6.  Comparison^of  Time  and  Longitude. 

7.  Miscellaneous  Tables. 


Dry. 


8.  Metric  System. 


1.  Historical. 


2.  Terms.. 


3.  Units 


1.  How  Derived.  f  1.  Milli. 

2.  Lower  Denominations...  -j    2.  Centi. 

I  3.  Deci. 

r  1.  Deka. 
Hekto. 

3.  Kilo. 

I  4.  Myria. 

1.  Meter. 

2.  Ar. 

3.  Liter. 


3.  Higher  Denominations.. 

1 


I  4   Gram. 

11.  Length. 
2.  Surface. 
3.  Capacity. 
4.  Weight. 


5.  Principles. 

6.  Table  of  Legal  and  Approximate  Values. 


154  RAY'S  HIQHER  ARITHMETIC. 


KEDUCTION  OF  COMPOUND  NUMBEKS. 

218.  Reduction  of  Compound  Numbers  is  the  process 
of  changing  them  to  equivalent  numbers  of  a  different  de- 
nomination. 

Reduction  takes  place  in  two  ways : 
From  a  higher  denomination  to  a  lower. 
From  a  lower  denomination  to  a  higher. 

PRINCIPLES. — 1.  Reduction  from  a  higher  denomination  to 
a  lower,  is  performed  by  multiplication. 

2.  Reduction  from  a  lower  denomination  to  a  higher,  is  per- 
formed by  division. 

PROBLEM.  — Reduce  18  bushels  to  pints. 

OPERATION. 

SOLUTION. — Since  1  bu.  —  4  pk.,  18  bu.  1  8  bu 

=  18  times  4  pk.  =  72  pk.,  and  since  1  pk.  4 

=  8  qt.,  72  pk.  =  72  times  8  qt.  —  576  qt.;  y^     k 

and  since  1  qt.  =  2  pt.,  576  qt.  =  576  times  g 

2  pt.  =  1152  pt.     Or,  since   1   bu.  =  64  pt.,  ~57~6     t 

multiply  64  pt.  by  18,  which  gives  1152  pt.  2 
as  before.     This  is  sometimes  called  Beduc- 

i.        r>  *.  1  1  52  pt. 

tion  Descending. 

18  bu.  =  1 1  5  2  pt. 

PROBLEM. — Reduce  236  inches  to  yards. 

SOLUTION. — Since   12   inches  =  1    ft.,  OPERATION. 

236  inches  will  be  as  many  feet  as  12  in.  12)236  in. 

is  contained  times  in  2o6  in.,  which  is  3  )     1  9  2  ft. 

19|  ft.,  and  since  3  ft.  =  1  yd.,  19|  ft  will  ~~^T  yd. 

be  as  many  yd.  as  3  ft.  is  contained  times  236  in  —  6  ^  yd 
in  19f  ft.,  which   is  6f  yd.     Or,  since  1 

yd.  =  36  in.,  divide  236  in.  by  36  in.,  which  gives  6f  yd.,  as  before. 
This  is  sometimes  called  Reduction  Ascending. 

NOTE. — In  the  last  example,  instead  of  dividing  236  in.  by  36  in. 
the  unit  of  value  of  yards,  since  1  inch  is  equal  to  -£•$  yards,  236 
inches  =  236  X  sV  =  W  yd.  —  6f  yd.  The  operation  by  division 
is  generally  more  convenient. 


REDUCTION  OF  COMPOUND  NUMBERS.       155 

REMARK. — Reduction  Descending  diminishes  the  size,  and,  there- 
fore, increases  the  number  of  units  given ;  while  Reduction  Ascend- 
ing increases  the  size,  and,  therefore,  diminishes  the  number  of  units 
given.  This  is  further  evident  from  the  fact,  that  the  multipliers 
in  Reduction  Descending  are  larger  than  1 ;  but  in  Reduction  As- 
cending smaller  than  1. 

PROBLEM. — Reduce  f  gallons  to  pints. 

SOLUTION.— Multiply  by  4  to  OPERATION. 

reduce  gal.  to  qt. ;  then  by  2  to  _3  X  £  X  $  =  3  pt. 

reduce  qt.    to  pt.      Indicate  the  jg 

operation,  and  cancel.  |.  gal.  =  3  pt. 

PROBLEM. — Reduce  5-f-  gr.  to  §. 

OPERATION. 

SOLUTION.  —  Al-  2 

though  this  is  Re-  _A^  y_JL  ylv— =-—    Z 
duction  Ascending,             ^  T  Sr*         j         2038        84 
we  use  Prin.  1,  in  . 

multiplying  by  the  successive 
unit     values,     •£-$,     J,    and    J.  ^  T  £>r<        "STo- 

PROBLEM. — Reduce  9.375  acres  to  square  rods. 

OPERATION. 

9.375 
160 


562500 
9375 

1500.000  sq.  rd. 
9.3  7  5  A.  —  1  5  0  0  sq.  rd. 

PROBLEM. — Reduce  2000  seconds  to  hours. 

OPERATION. 

2000X^X;^  =  ~   =|   hr. 
6  ft       b  ft       oo       9 

2000  sec.  =       hr. 


156  RA  Y>  S  HIGHER  ARITHMETIC. 

PROBLEM. — Keduce  1238.73  hektograms  to  grams. 

OPERATION. 

123  8.7  3X100  =  123873  grams. 


PROBLEM. — How  many  yards  in  880  meters? 

RATION. 

-=  9  6  2.3  7  7 -f  yd. 


OPERATION. 

3  9.3  7  in.  X  8  8  0 


12X3 

REMARK. — Abstract  factors  can  not  produce  a  concrete  result; 
sometimes,  however,  in  the  steps  of  an  indicated  solution,  where  the 
change  of  denomination  is  very  obvious,  the  abbreviations  may  be 
omitted  until  the  result  is  written. 

From  the  preceding  exercises,  the  following  rules  are 
derived : 

219.  For  reducing  from  higher  to  lower  denomina- 
tions. 

Rule. — 1.  Multiply  the  higJiest  denomination  given,  by  that 
number  of  the  next  lower  which  makes  a  unit  of  the  higher. 

2.  Add    to   the  product   the   number,   if  any,    of  the   lower 
denomination. 

3.  Proceed   in  like   manner   with   the   result   thus   obtained, 
till  the  whole  is  reduced  to  the  required  denomination. 

220.  For  reducing  from  lower  to  higher  denomina- 
tions. 

Rule. — 1.  Divide  the  given  quantity  by  that  number  of  its 
own  denomination  which  makes  a  unit  of  the  next  higher. 

2.  Proceed   in    like   manner  with  the  quotient  thus  obtained, 
till  the  whole  is  reduced  to  the  required  denomination. 

3.  The    last   quotient,    with   the  several   remainders,  if  any, 
annexed,  will  be  the  answer. 

NOTE. — In  the  Metric  System  the  operations  are  performed  by 
removing  the  point  to  the  right  or  to  the  left. 


REDUCTION  OF  COMPOUND  NUMBERS.       157 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  square  rods  in  a  rectangular  field  18.22 
chains  long  by  4.76  ch.  wide?  1387.6352  sq.  rd. 

2.  Reduce  16.02  chains  to  miles.  .20025  mi. 

3.  How  many  bushels  of  wheat  would  it  take  to.  fill  750 
hektoliters?  2128J  bu. 

4.  Eeduce  35.781  sq.  yd.  to  sq.  in.         46372.176  sq.  in. 

5.  Reduce  10240  sq.  rd.  to  sq.  ch.  640  sq.  ch. 

6.  How    many    perches    of   masonry    in    a    rectangular 
solid   wall   40   ft.  long  by  7-|  ft.  high,  and   2f  ft.  average 
thickness?  32ff  P. 

7.  How  many  ounces  troy  in    the    Brazilian  Emperor's 
diamond,  which  weighs  1680  carats?  11.088  oz. 

8.  Reduce  75  pwt.  to  3.  30  3. 

9.  Reduce  $  gr.  to  §.  -^  §. 

10.  Reduce  18f  3  to  oz.  av.  2-f-  oz. 

11.  Reduce  96  oz.  av.  to  oz.  troy.  87^  oz.  troy. 

12.  How  many  gal.   in  a  tank  3  ft.  long  by  2^  ft.  wide 
and  \\  ft.  deep?  75ff  gal. 

13.  How  many  bushels  in   a  bin  9.3  ft.  long  by  3-f  ft, 
wide  and  2^  ft.  deep?  61  bu.,  nearly. 

14.  How  many  sters  in  75  cords  of  wood?      271.837+  s. 

15.  Reduce  2J  years  to  seconds.  70956000  sec. 

16.  Forty-nine  hours  is  what  part  of  a  week?  -^  wk. 

17.  Reduce  90.12  kiloliters  to  liters.  90120  1. 

18.  Reduce  25"  to  the  decimal  of  a  degree.  .00694° 

19.  Reduce  192  sq.  in.  to  sq.  yd.  2T  S(l-  J^- 

20.  Reduce  6|  cu.  yd.  to  cu.  in.  311040  cu.  in. 

21.  Reduce  $117.14  to  mills.  117140  mills. 

22.  Reduce  6.19  cents  to  dollars.  $.0619 

23.  Reduce  1600  mills  to  dollars.  $1.60 

24.  Reduce  $5f  to  mills.  5375  mills. 

25.  Reduce  12  Ib.  av.  to  Ib.  troy.  14T7T  lb. 

26.  How  many  grams  in  6.45  quintals?  645000  g. 

27.  Reduce  .216  gr.  to  oz.  troy.  .00045  oz.  troy. 


158  RAY'S  HIGHER  ARITHMETIC. 

28.  Reduce  47.3084  sq.  mi.  to  sq.  rd.       4844380.16  sq.  rd. 

29.  Reduce  4^  B  to  ft.  fa  ft. 

30.  Reduce  7-J-  oz.  av.  to  cwt.  -^-g-  cwt. 

31.  Reduce  99  yd.  to  miles.  T|^  mi. 

32.  How  many  acres   in   a   rectangle    24±   rd.    long    by 
16.02  rd.  wide?  2.4530625  acres. 

33.  How  many  cubic  yards  in  a  box  6^  ft.  long  by  2^  ft. 
wide  and  3  ft.  high?  l-f~|  cu.  yd. 

34.  Reduce  169  ars  to  square  meters.  16900  m2. 

35.  Reduce  2£  f§  to  r^.  1200  n^. 

36.  If  a  piece  of  gold  is  y  pure,  how  many  carats  fine 
is  it?  20f  carats. 

37.  In  18|  carat  gold,  what  part  is  pure  and  what  part 
alloy?  ff  pure,  and  ^  alloy. 

38.  How  many  square  meters  of  matting  are  required  to 
cover  a  floor,  the  dimensions  of  which   are  6   m.,  1^   dm. 
by  5  m.,  3  cm.?  30.9345  m2. 

39.  How    many   cords    of  wood  in    a    pile  120  ft.  long, 
6£  ft.  wide,  and  8f  ft.  high  ?  53TVg-  C. 

40.  How  many  sq.  ft.  in  the  four  sides  of  a  room  21|-  ft. 
long,  16i  ft.  wide,  and  13  ft.  high?  988  sq.  ft. 

41.  What  will  be  the  cost  of  27  T.  18  cwt.  3  qr,  15  Ib. 
12  oz.  of  potash,  at  $48.20  a  ton?  $1346.97—. 

42.  What  is  the  value  of  a  pile   of  wood  16  m.,  1  dm., 
5  cm.  long,  1  m.,  2   dm.,   2   cm.   wide,  and  1  m.,   6   dm., 
8  cm.  high,  at  $2.30  a  ster?  $76.13+ 

43.  What  is  the  cost  of  a  field  173  rods  long  and  84  rods 
wide,  at  $25.60  an  acre?  $2325.12 

44.  If  an   open   court   contain    160   sq.   rd.    85   sq.    in. ; 
how  many  stones,  each  5  inches  square,  will  be  required  to 
pave  it?  250909  stones. 

45.  A   lady  had    a    grass-plot    20    meters    long    and    15 
meters    wide ;     after    reserving    two    plots,    one    2    meters 
square  and  the   other    3   meters  square,  she   paid  51  cents 
a  square  meter  to  have  it  paved  with  stones :  what  did  the 
paving  cost?  $146.37 


REDUCTION  OF  COMPOUND  NUMBERS.       159 

46.  A  cubic  yard  of  lead  weighs  19,128  lb.:  what  is  the 
weight  of  a  block  5  ft.  3^  in.  long,  3  ft.  2  in.  wide,  and 
1  ft.  8  in.  thick?  9  T.  17  cwt.  7  lb.   11.37  oz. 

47.  A  lady  bought  a  dozen  silver  spoons,  weighing  3  oz. 
4  pwt.   9  gr.,  at  $2.20  an  oz.,  and  a   gold  chain  weighing 
13  pwt.,  at  $1J  a  pwt.:  required  the  total  cost  of  the  spoons 
and  chain.  $23.331  ' 

48.  A  wagon-bed  is  10|-  ft.  long,  3|  ft.  wide,  and  1|  ft. 
deep,  inside  measure :  how  many  bushels  of  corn  will  it  hold, 
deducting  one  half  for  cobs?  22  bu.  4  qt.  1.5 —  pt. 

49.  If  a  man  weigh  160    lb.    avoirdupois,  what  will  he 
weigh  by  troy  weight  ?  194  lb.  5  oz.  6  pwt.  16  gr. 

50.  The  fore -wheel  of  a  wagon  is  13  ft.  6  in.  in  circum- 
ference, and  the  hind  wheel  18  ft.  4  in.:  how  many  more 
revolutions   will  the  fore-wheel  make  than  the  hind  one  in 
50  miles?  5155.55+  revolutions. 

51.  An    apothecary  bought    5  lb.    10  §.    of  quinine,    at 
$2.20  an  ounce,  and  sold  it  in  doses  of  9  gr.,  at  10  cents 
a  dose:  how  much  did  he  gain?  $219.33| 

52.  How  many  steps  must  a  man  take  in  walking  from 
Kansas  City  to  St.  Louis,  if  the  distance  be  275  miles,  and 
each  step,  2  ft.  9  in.?  528000  steps. 

53.  The  area  of  Missouri    is    65350  sq.  mi.:    how  many 
hektars  does  it  contain?  16925940.91+  Ha. 

54.  A  school-room   is    36  ft.  long,   24  ft.  wide,   and    14 
ft.  high ;    required   the    number    of  gallons    of  air   it  will 
contain?  90484.36+  gal. 

55.  Allowing  8   shingles    to  the    square  foot,  how  many 
shingles    will    be   required    to    cover    the    roof   of  a    barn 
which   is   60  feet  long,  and   15  feet  from  the  comb  to  the 
eaves  ?  14400  shingles. 

56.  A   boy  goes    to    bed  30  minutes    later,  and  gets  up 
40  minutes    earlier  than    his   room-mate :    how  much   time 
does    he    gain    over    his    room-mate    for    work    and    study 
in    the    two    years    1884    and    1885,    deducting    Sundays 
only?  731  hours,  30  min. 


160  RAY'S  HIGHER  ARITHMETIC. 


ADDITION  OF  COMPOUND  NUMBERS. 

221.  Compound  Numbers  may  be  added,  subtracted, 
multiplied,  and  divided.  The  principles  upon  which  these 
operations  are  performed  are  the  same  as  in  Simple  Num- 
bers, with  this  variation ;  namely,  that  in  Simple  Numbers 
ten  units  of  a  lower  denomination  make  one  of  the  next 
higher,  while  in  Compound  Numbers  the  scales  vary. 

Addition  of  Compound  Numbers  is  the  process  of  find- 
ing the  sum  of  two  or  more  similar  Compound  Numbers. 

PROBLEM. — Add  3  bu.  2J  pk.;  1  pk.  1^  pt.;  5  qt.  1  pt.; 
2  bu.  1£  qt;  and  .125  pt. 

SOLUTION. — Reduce  the   frac-  OPERATION. 

tion   in   each   number   to   lower        bu.  pk.   qt.   pt. 
denominations,  and  write   units          3      2       2      0     =  3  bu.  2J  pk. 
of  the  same  kind   in  the   same  1       0       1 1  =  1  pk.  1  \  pt. 

column.   The  right-hand  column,  5       1     =  5  qt.  1    pt. 

when  added,  gives  3^-  pt.  =  1  qt.  20  1  f  =  2  bu.  1  \  qt. 
l^j  pt.;  write  the  l/^  and  add  J— .125  pt. 

the  1  qt.  with  the  next  column,          601       1-^=  Ans. 
making  9  qt.  =  1  pk.  1  qt.;  write 

the  1  qt.  and  carry  the  1  pk.  to  the  next  column,  making  4  pk.  =  1  bu.; 
as  there  are  no  pk.  left,  write  down  a  cipher  and  carry  1  bu.  to  the 
next  column,  making  6  bu. 

PROBLEM.— Add  2  rd.  9  ft.  TI  in.;  13  ft.  5.78  in.;  4  rd. 
11  ft.  6  in.;  1  rd.  lOf  ft;  6  rd.  14  ft.  6f  in. 

OPERATION. 

SOLUTION. — The  numbers  are  prepared,  rd.  ft.  in. 

written,  and   added,  as   in   the   last   ex-  2  9  7.25 

ample;  the  answer  is  16  rd.  9|  ft.  9.655  13  5.78 

in.      The   J   foot   is   then   reduced   to    6  4  11  6 

inches,  and  added  to  the  9.655  in.,  making  1  10  8 

15.655   in.  =  l   ft.  3.655   in.      Write   the  6  14  6.625 

3.655  in.,  and  carry  the  1  ft.,  which  gives  16  9J  9.655 
16  rd.  10  ft.  3.655  in.  for  the  final  answer.  but  £  ft.  =  6. 

16  10  3.655 


ADDITION  OF  COMPOUND  NUMBERS.        161 

Rule. — 1.    Write  the  members  to  be  added,  placing  units  of 
the  same  denomination  in  the  same  column. 

2.  Begin   with   the    lowest    denomination,  add   the   numbers, 
and   divide   their  sum   by  the  number  of  units  of  this   denom- 
ination whidi  make  a  unit  of  the  next  higher. 

3.  Write  the  remainder  under  the  column  added,  and  carry 
the  quotient  to  the  next  column. 

4.  Proceed  in  the  same  manner  with  all   the  columns  to  the 
last,  under  which  write  its  entire  sum. 

REMARK. — The  proof  of  each   fundamental  operation  in  Com- 
pound Numbers  is  the  same  as  in  Simple  Numbers. 

EXAMPLES  FOR  PRACTICE. 

1.  Add  %  mi.;  146 J-  rd.;  10  mi.  14  rd.  7  ft.  6  in.;  209.6 
rd.;  37  rd.   16  ft.  2^  in.;  1  mi.  12  ft.  8.726  in. 

12  mi.  180  rd.  9  ft.  4.633|  in. 

2.  Add  6.19  yd.;  2  yd.  2  ft.  9f  in.;  1  ft.  4.54  in.;  10  yd. 
2.376  ft.;  f  yd.;  If  ft.;  |  in.  21  yd.  2  ft.  3.517  in. 

3.  Add  3  yd.  2  qr.  3  na.  1|  in.;   1  qr.  2|  na.;  6  yd.  1  na. 
2.175  in.;   1.63  yd.;  f  qr.;  f  na,  12yd.  1  na.  0.755  in. 

4.  If  the  volume  of  the  earth  is  1;  Mercury,  .06;  Venus, 
.957;  Mars,  .14;  Jupiter,  1414.2;  Saturn,  734.8;  Uranus, 
82;  Neptune,  110.6;    the  Sun,    1407124;    and    the  Moon, 
.018,  what  is  the  volume  of  all?  1409467.775 

5.  James  bought  a  balloon  for  9  francs  and  76  centimes, 
a  ball  for  68  centimes,  a  hoop  for  one  franc   and  37  cen- 
times, and  gave  to  the  poor  2  francs  and  65  centimes,  and 
had  3  francs  and  4  centimes   left.     How  much  money  did 
he  have  at  first?  17^  francs. 

6.  Add  15  sq.  yd.  5  sq.  ft.  87  sq.  in.;   16 J  sq.  yd.;  10  sq. 
yd.  7.22  sq.  ft.;  4  sq.  ft.  121.6  sq.  in.;  ^  sq.  yd. 

43  sq.  yd.  7  sq.  ft.  37.78  sq.  in. 

7.  Add  101  A.  98.35  sq.  rd.;  66  A.  74J-  sq.  rd.;  20  A.;  12 
A.  113  sq.  rd.;  5  A.  13.33-*  sq.  rd.         205  A.  139.18£sq.  rd. 

H.  A.  14. 


162  RAY'S  HIGHER  ARITHMETIC. 

8.  Add  23  cu  yd.  14  cu.  ft.  1216  cu.  in.;  41  cu.  yd.  6 
cu.  ft.  642.132  cu.  in.;  9  cu.  yd.  25.065  cu.  ft.;  ^  cu.  yd. 

75  cu.  yd.  4  cu.  ft.  1279.252  cu.  in. 

9.  Add  f  C.;  f  cu.  ft.;  1000  cu.  in. 

107  cu.  ft.  1072  cu.  in. 

10.  Add  2  lb.  troy,  6|  oz.;  If  lb.;  12.68  pwt.;  11  oz.  13 
pwt.  19i  gr.;  |  lb.  -if  oz.;  f  pwt. 

5  lb.  troy,  9  oz.  9  pwt.  2.85-J-  gr. 

11.  Add8g  14.6  gr.;  4.18  g;  7T\  3 ;  23  29  18  gr.;  Ig 
12  gr.;  i  9.  1  ft.  2  g  4  3  1  9- 

12.  Add  T5e  T.;  9  cwt.  1  qr.  22  lb.;  3.06  qr.;  4  T.  8.764 
cwt.;  3  qr.  6  lb.;  -^  cwt.  5  T.  6  cwt.  2  qr.  14^.  lb. 

13.  Add  .3  lb.  av.;  f  oz.  5Jf  oz. 

14.  Add  6  gal.  3£  qt.;  2  gal.  1  qt.   .83  pt.  ;  1  gal.  2  qt. 
|  pt.;  |  gal.;   g  qt.;  $  pt.  11  gal.  2  qt.  .llfj-  pt. 

15.  Add  4  gal.  .75  pt.;  10  gal.  3  qt.  1£  pt.;  8  gal.  f  pt.; 
5.64  gal.;  2.3  qt.;  1.27  pt.;  ^  pt-         29  gal.  2  qt.  ,05f  pt. 

16.  Add  1  bu.  %  pk.;  T6T  bu.;  3  pk.  5  qt.   l\  pt.;  9  bu. 
3.28  pk.;  7  qt.  1.16  pt.;  -f\  pk.  12  bu.  3  pk.  .46^1  pt. 

17.  Add  |  bu.;  |  pk.;  f  qt.;  f  pt.  2  pk.  ff  pt. 

18.  Add  6  f5  2  fg  25  nt ;  2J  f§;  7  ft  42  nt  ;  1  f  §  2|  fs; 
3  fa  6fs51  nt.  14  fg  7  %  38  nt. 

19.  Add  -J-  wk.;  \  da.;  4^  lir.;  \  min. ;  -|-  sec. 

4  da.  30  min.  301  sec. 

20.  Add  3.26  yr.  (365  da.  each);  118  da.  5  hr.  42  min. 
37|-  sec.;  63.4  da.;  7|  hr.;  1  yr.  62  da.  19  hr.  24f  min.; 
T^32-  da.  4  yr.  340  da,  1  hr.  14  min.  55|-  sec. 

21.  Add  27°  14'  55.24";  9°  18 J";  1°  15f ;  116°  44' 
23.8"  154°  14'  57.29" 

22.  Add  $i;  |  ct.;  -|  m.  50  ct.  2f  m. 

23.  Add  3  dollars  7  m.;  5  dollars  20  ct.;  100  dollars  2  ct. 
6  m.;  19  dollars  \  ct.  $127  23  ct.  4J  m. 

24.  Add    £21    6s.  3Jd.;    £5   17|s.;    £9.085;    16s.  7Jd.; 

£37  10s.  8.15d. 

25.  Add  |  A.;  f  sq.  rd.;  i  sq.  ft. 

107  sq.  rd.  12  sq.  yd.  6  sq.  ft.  344-  sq.  in. 


SUBTRACTION  OF  COMPOUND  NUMBERS.     163 


SUBTKACTION  OF  COMPOUND  NUMBEKS. 

222.  Subtraction  of  Compound  Numbers  is  the 
process  of  finding  the  difference  between  two  similar  Com- 
pound Numbers. 

PROBLEM. --From  9  yd.  1  ft.  6^  in.  take  1  yd.  2.45  ft. 

SOLUTION.— Change  the  J  in.  to  a  decimal,  OPERATION. 
making  the  minuend  9  yd.  1  ft.  6.5  in.;  reduce  yd.  ft.  in. 
.45  ft.  to  inches,  making  the  subtrahend  1  yd.  2  9  1  6.5 
ft.  5.4  in.  The  first  term  of  the  difference  is  1.1  1  2  5.4 

in.     To  subtract  the  2  ft.,  increase  the  minuend  7     2     1.1  Ans. 

term  by  3  feet,  and  the  next  term  of  the  subtra- 
hend by  the  equivalent,  1  yard.     Taking  2  ft.  from  4  ft.  we  have  a 
remainder  2  ft.,  and  2  yd.  from  9  yd.  leaves  7  yd.,  making  the  answer 
7  yd.  2  ft.  1.1  in. 

PROBLEM. — From  2  sq.  rd.  1  sq.  ft.  take  1  sq.  rd.  30  sq. 
yd.  2  sq.  ft. 

SOLUTION. — Write  the  numbers  as  before.  OPERATION. 

If  the  1  sq.ft.be    increased  by  a  whole  sq.     sq.rd.  sq.yd.  sq.ft. 
yd.  and  the  next  higher  part  of  the  subtra-          201 

hend  by  the  same  amount,  we  shall  have  to          1         30 2 

make  an  inconvenient  reduction  of  the  first  Ans.  1J  sq.ft. 

remainder  as  itself  a  minuend.     Hence,  we 

make  the  convenient  addition  of  J  sq.  yd.,  and,  subtracting  2  sq.  ft. 
from  3J  sq.  ft.,  we  have  1 J-  sq.  ft.  Then,  giving  the  same  increase  to 
the  30  sq.  yd.,  we  proceed  as  in  the  former  case,  increasing  the  upper 
by  one  of  the  next  higher,  and,  having  no  remainder  higher  than 
feet,  the  answer  is,  simply,  1}  sq.  ft.  =  1  sq.  ft.  36  sq.  in. 

Rule. — 1.  Place  the  subtrahend  under  the  minuend,  so  that 
numbers  of  the  same  denomination  stand  in  the  same  column. 
Begin  at  the  lowest  denomination,  and,  subtracting  the  parts 
successively  from  right  to  left,  write  the  remainders  beneath. 

2.  If  any  number  in  the  subtrahend  be  greater  than  that 
of  the  same  denomination  in  the  minuend,  increase  the  upper 
by  a  unit,  or  such  other  quantity  of  the  next  higher  denomina- 


164  RAY'S  HIGHER  ARITHMETIC. 

tion  as  will  render  the  subtraction  possible,  and  give  an  equal 
increase  to  the  next  higher  term  of  the  subtrahend. 

KEMARK. — The  increase  required  at  any  part  of  the  minuend  is, 
commonly,  a  unit  of  the  next  higher  denomination.  In  a  few 
instances  it  will  be  convenient  to  use  more,  and,  if  less  be  required, 
the  tables  will  show  what  fraction  is  most  convenient.  Sometimes 
it  is  an  advantage  to  alter  the  form  of  one  of  the  given  quantities 
before  subtracting. 


EXAMPLES  FOR  PRACTICE. 

1.  Subtract  f  mi.  from  144.86  rd.  16.86  rd. 

2.  Subtract  1.35  yd.  from  4  yd.  2  qr.   1  na.   If  in. 

3  yd.  1  qr.  f  in. 

3.  Subtract  2  sq.  rd.  24  sq.  yd.  91  sq.  in.  from  5  sq.  rd. 
16  sq.  yd.  6§  sq.  ft.       2  sq.  rd.  22  sq.  yd.  8  sq.  ft.  41  sq.  in. 

4.  Subtract  384  A.  43.92  sq.  rd.  from  1.305  sq.  mi. 

450  A.  148.08  sq.  rd. 

5.  Subtract  13  cu.  yd.  25  cu.  ft.  1204.9  cu.  in.  from  20  cu. 
yd.  4  cu.  ft.  1000  cu.  in.          6  cu.  yd.  5  cu.  ft.  1523.1  cu.  in. 

6.  Subtract  9.362  oz.  troy  from  1  lb.  15  pwt.  4  gr. 

3  oz.  7  pwt.  22.24  gr. 

7.  Subtract  f£  3  from  T6T  g.  3  3  1  9  15^  gr. 

8.  Subtract  56  T.  9  cwt.  1  qr.  23  lb.  from  75.004  T. 

18  T.  10  cwt.  2  qr.  10  lb. 

9.  Subtract  T\  lb.  troy  from  ^  lb.  avoirdupois.  0. 

10.  Subtract  12  gal.  1  qt.  3  gills  from  31  gal.  1|  pt. 

18  gal.  3  qt.  3  gi. 

11.  Subtract  .0625  bu.  from  3  pk.  5  qt.  1  pt. 

3  pk.  3  qt.  1  pt. 

12.  Subtract  1  f§  4  fg  38  nt  from  4  f g  2  fg. 

2"f§  5fe  22  HL. 

13.  Subtract  275  da.  9  hr.  12  min.  59  sec.  from  2.4816  yr. 
(allowing  365^  days  to  the  year.) 

1  yr.  265  da.  18  hr.  29  min.  21.16  sec. 


MULTIPLICATION  OF  COMPO  UND  NUMBERS.  165 

14.  Find  the  difference  of  time  between  Sept.  22d,  1855, 
and  July  1st,  1856.  9  mon.  9  da. 

15.  Find  the  difference  of  time  between  December  31st, 
1814,  and  April  1st,  1822.  7  yr.  3  mon. 

16.  Subtract  43°  18'  57.18"  from  a  quadrant. 

46°  41'  2.82" 

17.  Subtract  161°  34'  11.8"  from  180°.         18°  25'  48.2" 

18.  Subtract  ££  ct,  from  $^\.  8  ct.  4|  m. 

19.  Subtract  5  dollars  43  ct.  2^  m;  from  12  dollars  6  ct. 
8J-  m.  $6.635f 

20.  Subtract  £9  18s.  G^d.  from  £20.  £10  Is.  5|d. 

21.  From  %  A.  10  sq.  in.  take  79  sq.  rd.  30  sq.  yd.  2  sq. 
ft.  30  sq.  in.  16  sq.  in. 

22.  From  3  sq.  rd.  1  sq.  ft.  1  sq.  in.  take  1  sq.  rd.  30  sq. 
yd.  1  sq.  ft.  140  sq.  in.  1  sq.  rd.  1  sq.  ft.  41  sq.  in. 

23.  From  3  rd.  2  in.  take  2  rd.  5  yd.  1  ft.  4  in.         4  in. 

24.  From  7  mi.  1   in.  take  4  mi.  319  rd.  16  ft.  3  in. 

2  mi.  4  in. 

25.  From  13  A.  3  sq.  rd.  5  sq.  ft.  take  11  A.  30  sq.  yd.  8  sq 
ft.  40  sq.  in.  2  A.  1  sq.  rd.  30  sq.  yd.  1  sq.  ft.  32  sq.  in. 

26.  From  18  A.  3  sq.  ft.  3  sq.  in.  take  15  A.  3  sq.  rd.  30  sq. 
yd.  1  sq.  ft.  142  sq.  in.  2  A.  156  sq.  rd.  3  sq.  ft.  41  sq.  in. 


MULTIPLICATION  OF  COMPOUND  NUMBEKS. 

223.     Compound  Multiplication  is  the  process  of  mul- 
tiplying a  Compound  Number  by  an  Abstract  Number. 

PROBLEM. — Multiply  9  hr.  14  min.  8.17  sec.  by  10. 

OPERATION. 

SOLUTION. — Ten    times    8.17    sec.  = 

81.7  sec.  =  1  min.  21.7  sec.     Write  21,7  da"     ^^     ^ 

sec.   and  carry  1   min.  to  the  140  min. 

obtained    by  the    next   multiplication.  — 

This    gives    141    min.  =  2  hr.  21  min. 

Write  21   min.   and  carry  2  hr.     This  gives  92  hr.  =  3  da.  20  hr. 


166  RA  F>#  HIGHER  ARITHMETIC. 

PROBLEM.  —Multiply  12  A.  148  sq.  rd.  28f  sq.  yd.  by  84. 

SOLUTION.—  Since   84  =  7  X  12,   multiply  OPERATION. 

by  one  of   these    factors,  and    this    product  A.   sq.  rd.  sq.  yd. 

by  the  other  ;    the  last   product    is   the  one 

required.     The  same  result  can  be  obtained  __  7 
by  multiplying  by  84  at  once  ;    performing  90        82      1  7  \ 

the  work  separately,  at  one  side,  and  trans-  _  1  2 

ferring  the  results.  1086        30      24 

Rule.  —  1.    Write  the  multiplier  under  the   lowest  denomina- 
tion of  the  multiplicand. 

2.  Multiply    the   lowest    denomination  first,    and    divide   the 
product   by  the   number  of  units   of  this   denomination    which 
make  a  unit  of  the  next  higher;   ivrite  the  remainder  under  the 
denomination  multiplied,  and  carry  the  quotient  to  the  product 
of  the  next  higher  denomination. 

3.  Proceed  in  like  manner  with  all  the  denominations,  writing 
the  entire  product  at  the  last. 


EXAMPLES  FOR  PRACTICE. 

1.  Multiply  7  rd.  10  ft.  5  in.  by  6.  45  rd.  13  ft. 

2.  Multiply  1  mi.  14  rd.  8^  ft.  by  97. 

101  mi.  126  rd.  8  ft.  3  in. 

3.  Multiply  5  sq.  yd.  8  sq.  ft.  106  sq.  in.  by  13. 

77  sq.  yd.  5  sq.  ft.  82  sq.  in. 

4.  Multiply  41  A.  146.1087  sq.  rd.  by  9.046 

379  A.  23.4593+  sq.  rd. 

5.  Multiply  10  cu.  yd.  3  cu.  ft.  428.15  cu.  in.  by  67. 

678  cu.  yd.  1  cu.  ft.  1038.05  cu.  in. 

6.  Multiply  7  oz.  16  pwt.  5|  gr.  by  174. 

113  Ib.  3  oz.  5  pwt.  16|  gr. 

7.  Multiply  2  3  1  B  13  gr.  by  20.  6  §  3  3. 

8.  Multiply  16  cwt.  1  qr.  7.88  Ib.  by  11. 

8  T.  19  cwt.  2  qr.  11.68  Ib. 


DIVISION  OF  COMPOUND  NUMBERS.         167 

9.  Multiply  5  gal.  3  qt.  1  pt.  2  gills  by  35.108 

208  gal.   1  qt.   1  pt.  2.52  gills. 

10.  Multiply  26  bu.  2  pk.  7  qt.  .37  pt.  by  10. 

267  bu.  7  qt.  1.7  pt. 

11.  Multiply  3  fg  48  r^  by  12.  5  f§  5  %  36  n^ 

12.  Multiply  18  da.  9  hr.  42  min.  29.3  sec.  by  16T7T. 

306  da.  4  hr.  25  min.  2  sec.,  nearly. 

13.  Multiply  £215  16s.  2£d.  by  75.         £16185  14s.  |d. 

14.  Multiply  10°  28'  42f '  by  2.754        28°  51'  27.765" 


DIVISION  OF  COMPOUND  NUMBEKS. 

224.  Division  of  Compound  Numbers  is  the  process 
of  dividing  when  the  dividend  is  a  Compound  Number. 
The  divisor  may  be  Simple  or  Compound,  hence  there 
are  two  cases : 

1.  To  divide  a  Compound  Number  into  a  number  of  equal 
parts. 

2.  To  divide  one  Compound  Number  by  another  of  the  same 
kind. 

NOTE. — Problems  under  the  second  case  are  solved  by  reducing 
both  Compound  Numbers  to  the  same  denomination,  and  then 
dividing  as  in  simple  division. 

PROBLEM. — Divide  5  cwt.  3  qr.  24  Ib.  14f  oz.  of  sugar 
equally  among  4  men. 

SOLUTION. — 4  into   5    cwt.  gives    a  OPERATION. 

quotient  1  cwt.,  with  a  remainder  1  cwt.,  cwt.  qr.     Ib.         oz. 

==  4  qr.,  to  be  carried  to  3  qr.,  making  4)5       3       24       14f 

7  qr.;  4  into  7  qr.  gives  1  qr.,  with  3  qr.,  1       1       24       1  5  { J 

=?  75  Ib.,  to  be  carried  to  24  Ib.,  =  99  Ib.; 

4  into  99  Ib.  gives  24  Ib.,  with  3  Ib.,  =  48  oz.,  to  be  carried  to  14f  oz., 
making  62f  oz.;  4  into  62f  oz.  gives  15JJ  oz.,  and  the  operation  is 
complete. 


168  RAY'S  HIGHER  ARITHMETIC. 

PROBLEM.  —If  $42  purchase  67   bu.    2  pk.    5  qt.    If  pt. 

of  meal,  how  much  will  $1  purchase? 

OPERATION. 

SOLUTION. — Since   42  =  6  X  7?   divide  bu.     pk.    qt.     pt. 

first   by   one   of    these   factors,   and   the          6)67       2       5       If 
resulting    quotient    by   the    other ;     the         7)11       1       0       1  3 3 
last  quotient  will  be  the  one  required.  1231  -2  3- 

Rule. — 1.  Write  the  quantity  to  be  divided  in  the  order 
of  its  denominations }  beginning  with  the  highest;  place  the 
divisor  on  the  left. 

2.  Begin  with  the  highest  denomination,  divide  each  number 
separately,  and  write  the  quotient  beneath. 

3.  If  a  remainder  occur  after  any  division,  reduce  it  to  the 
next   lower   denomination,  and,  before   dividing,  add    to  it   the 
number  of  its  denomination. 


EXAMPLES  FOR  PRACTICE. 

1.  Divide  16  mi.  109  rd.  by  7.  2  mi.  107  rd. 

2.  Divide  37  rd.   14  ft.  11.28  in.  by  18. 

2rd.  1ft.  8. 96  in. 

3.  Divide  675  C.  114.66  cu.  ft.  by  83. 

8  C.  18.3453+  cu.  ft. 

4.  Divide  10  sq.  rd.  29  sq.  yd.  5  sq.  ft.  94  sq.  in.  by  17. 

19  sq.  yd.  4  sq.  ft.  119^-f-  sq.  in. 

5.  Divide  6  sq.  mi.  35  sq.  rd.  by  221      17Q  A.  108|  sq.  rd. 

6.  Divide  1245  cu.  yd.  24  cu.  ft.   1627  cu.  in.  by  11.303 

110  cu.  yd.  6  cu.  ft.  338.4+  cu.  in. 

7.  Divide  3  g  7  3  18  gr.  by  12.  2  3  1  9  16J  gr. 

8.  Divide  600  T.  7  cwt.  86  Ib.  by  29.06 

20  T.  13  cwt.  20  Ib.  14  oz.  12+  dr. 

9.  Divide  312  gal.  2  qt.  1  pt.  3.36  gills  by  72|. 

4  gal.  1  qt.  1.79+  gills. 
10.  Divide  19302  bu.  by  6.215 

3105  bu.  2  pk.  6  qt.  1.5+  pt. 


LONGITUDE  AND   TIME.  169 

11.  Divide  76  yr.  108  da.  2  hr.  38  min.  26.18  sec.  by  45. 

1  yr.  254  da.  27  min.  31.25—  sec. 

12.  Divide  152°  46'  2"  by  9.  16°  58'  26|". 

225.     Longitude  and  Time  give  rise  to  two  cases: 

1.  To  find    the    difference    of  longitude    between  two  places 
when  the  difference  of  time  is  given. 

2.  To  find  the  difference  of  time  when  their  longitudes  are  given. 

PROBLEM. — The  difference  of  time  between  two  places  is 
4  hr.  18  min.  26  sec.:  what  is  their  difference  of  longitude? 

SOLUTION. — Every  hour  of   time  corre- 

.  _0      ,   ,      J    .        ,  .  OPERATION. 

sponds  to  15   of  longitude ;  every  minute 

-•r/    -c  i        -X   j  hr.         mm.         sec. 

of  time  to  15   of  longitude  ;  every  second 

of    time   to  15"  of  longitude  (Art.  207).  .  r 

Hence,    multiplying    the    hours    in    the 


64°        36'        30" 
difference    of   time   by  15   will   give   the 

degrees    in   the    difference    of    longitude, 

multiplying  the  minutes  of  time  by  15  will  give  minutes  (/)  of 
longitude,  and  multiplying  the  seconds  of  time  by  15  will  give 
seconds  (  "  )  of  longitude. 

PROBLEM. — The    difference    of    longitude    between    two 
places  is  81°  39'  22":  what  is  their  difference  of  time? 

SOLUTION.— 15    into    81° 

gives  5  (marked  hr.),  and  6°  OPERATION. 
to   be   carried.      Instead  of  15)81°            39'            22" 
multiplying   6  by  60,  add-  5  hr.         26  min.     3  7  TV  sec. 
ing  the  39',  and  then  divid- 
ing, proceed  thus :    15  into  6°    is   the  same  as  15  into  6  X  W  = 

—  =  24',  and   as  15  into  39X  gives  2X  for  a  quotient  and  9' 
15 

remainder,  the  whole  quotient  is  267  (marked  min.)j  and  remainder 


9^  9X60",  which,  divided  by  15,  gives  ==  36",  which 

15 

with  1T^,  obtained  by  dividing  22"  by  15,  gives  37T7/X  (marked  sec.). 
The  ordinary  mode  of  dividing  will  give  the  same  result,  and  may 
be  used  if  preferred. 


H.  A.  15- 


170 


' S  HIGHER  ARITHMETIC. 


226.  From  these  solutions  we  may  obtain  the  following 
rules : 

CASE    I. 

Rule. — Multiply  the  difference  of  time  by  15,  according  to 
the  rule  for  Multiplication  of  Compound  Numbers,  and  mark 
the  product  "  instead  of  hr.  min.  sec. 

CASE    II. 

Rule. — Divide  the  difference  of  longitude  %  15,  according 
to  the  rule  for  Division  of  Compound  Numbers,  and  mark  the 
quotient  hr.  min.  sec.,  instead  of  ''. 

NOTE. — The  following  table  of  longitudes,  as  given  in  the  records 
of  the  U.  S.  Coast  Survey,  is  to  be  used  for  reference  in  the  solution 
of  exercises.  "  W."  indicates  longitude  West,  and  "  E."  longitude 
East  of  the  meridian  of  Greenwich,  England. 

TABLE  OF  LONGITUDES. 


PLACE. 

LONGITUDE. 

PLACE. 

LONGITUDE. 

Portland,  Me.,  .... 

o       t        n 

70    15    18  W. 

Des  Moines  Iowa 

o       r        n 

93    37    16  W. 

Boston,  Mass.i  .... 

71      3    50  " 

Omaha,  Neb.,  

95    56    14   " 

New  Haven,  Conn.,  . 
New  York  City,  .  .  . 

72    55    45  " 
74      0    24 

Austin,  Tex.,  
Denver,  Col  

97    44    12  " 
104    59    33  " 

Philadelphia,  Pa.,  . 
Baltimore,  Md.,  .  .  . 
Washington,  D.  C.,  . 
Richmond,  Va.,  .  .  . 
Charleston,  S.  C  ,  .  . 
Pittsburgh,  Pa.,  ... 
Savannah,  Ga.,  .  .  . 
Detroit,  Mich.,  .  .  . 
Cincinnati,  O.,  ... 
Louisville  Ky 

75      9      3 
76    36    59 
77      0    36 

77    2G      4 
79    55    49 
80      2      0 
81      5    26 
83      3      0 
84    29    45 
85    25      0 

Salt  Lake  City,  Utah,  . 
San  Francisco,  Cal.,  .  . 
Sitka,  Alaska,  
St.  Helena  Island,  .  .  . 
Reykjavik,  Iceland,  .  . 
Rio  Janeiro,  Brazil,  .  . 
St.  Johns,  N.  F.,  .... 
Honolulu,  Sandwich  Is. 
Greenwich,  Eng.,  .... 
Paris  France 

111    53    47  " 
122    27    49   " 
135    19    42   " 
5    42      0   " 
22      0      0  " 
43    20      0  " 
52    43      0   " 
157    52      0   " 
000 
2    20      0   E 

Indianapolis  Ind 

86      6      0 

Rome  Italv.  ... 

12    28      0   " 

Nashville,  Tenn.,  .  . 
Chicago  111 

86    49      0 
87    35      0 

Berlin,  German  Em  p.,  . 
Vienna,  Austria 

13    23      0   " 
16    20      0   " 

Mobile,  Ala.,  .  .  . 
Madison,  Wis.,  .  .  . 
New  Orleans,  La.,  .  . 
St  Lonis  Mo 

88      2    28 
89    24      3 
90      3    28 
90    12    14 

Constantinople,Turkey, 
St.  Petersburg,  Russia,  . 
Bombay,  India,  
Pekin  China  

28    59      0   " 
30    16      0   " 
72    48      0   " 
116    96      0   " 

Minneapolis,  Minn., 

93    14      8 

Sydney,  Australia,  .  .  . 

151     11      0   " 

LONGITUDE  AND  TIME.  171 


EXAMPLES  FOR  PRACTICE. 

1.  It  is  six  o'clock  A.  M.  at  New  York ;  what  is  the  time 
at  Cincinnati?  18  min.  2.6  sec.  after  5  o'clock  A.  M. 

2.  The  difference  of  time   between   Springfield,  111.,  and 
Philadelphia  being  58  min.   1T2^  sec.,  what  is  the  longitude 
of  Springfield?  89°  39'  20"  W. 

3.  At  what  hour  must  a  man  start,  and  how  fast  would 
he  have  to  travel,  at  the  equator,  so  that  it  would  be  noon 
for  him  for  twenty-four  hours? 

Noon;  1037.4  statute  miles  per  hour. 

4.  What   is    the   relative   time  between  Mobile  and  Chi- 
cago? Chicago  time  1  min.  49|f  sec.  faster. 

5.  A  man  travels  from  Halifax  to  St.  Louis ;  on  arriving, 
his  watch  shows  9  A.  M.  Halifax  time.      The  time  in  St. 
Louis  being  13  min.  32T^  sec.  after  7  o'clock  A.  M.,  what 
is  the  longitude  of  Halifax?  63°  35'  18"  W. 

6.  Noon  occurs  46  min.  58   sec.   sooner  at  Detroit    than 
at  Galveston,  Texas :    what   is  the  longitude   of  the  latter 
place?  94°  47'  30"  W. 

7.  When  it  is  five  minutes  after  four  o'clock  on  Sunday 
morning  at  Honolulu,  what  is  the  hour  and  day  of  the  week 
at  Sydney,  Australia? 

41  min.  12  sec.  after  12  o'clock  A.  M.,  Monday. 

8.  What  is  the  difference  in  time  between  St.  Petersburg 
and  New  Orleans?   •  8  hr.  1  min.  17-j-f  sec. 

9.  When  it  is  one  o'clock   P.  M.  at  Rome,  it  is  54  min. 
34  sec.  after  6   o'clock  A.  M.  at    Buffalo,   N.  Y.:  what  is 
the  longitude  of  the  latter?  78°  53'  30"  W. 

10.  When    it   is   six   o'clock  P.  M.   at  St.  Helena,  what 
is  the  time  at  San  Francisco? 

12  min.  56||  sec.  after  10  o'clock  A.  M. 

11.  A  ship's  chronometer,  set  at  Greenwich,  points  to  4 
hr.  43  min.  12  sec.  P.  M.:  the  sun  being  on  the  meridian, 
what  is  the  ship's  longitude?  70°  48'  W. 


172 


RAY'S  HIGHER  ARITHMETIC. 


ALIQUOT  PAETS. 

227c     An  aliquot  part  is  an  exact  divisor  of  a  number. 

Aliquot  parts  may  be  used  to  advantage  in  finding  a 
product  when  either  or  when  each  of  the  factors  is  a 
Compound  Number. 

PROBLEM.— Find  the  cost  of  28  A.  145  sq.  rd.  15^  sq.  yd. 
of  land,  at  $16  per  acre. 

SOLUTION.— Multiply  $16,  the 
price  of  1  A.,  by  28 ;  the  product, 
$448,  is  the  price  of  28  A.  145 
sq.  rd.  is  made  up  of  120  sq.  rd., 
20  sq.  rd.,  and  5  sq.  rd.  120  sq. 
rd.  =  f  of  an  acre ;  hence  take  f 
of  $16,  the  price  per  acre,  to  find 
the  cost  of  120  sq.  rd.  20  sq.  rd. 
—  i  of  120  sq.  rd.,  and  cost  J  as 
much.  5  sq.  rd.  —  J  of  20  sq.  rd., 
and  cost  \  as  much.  15|  sq.  yd. 
=  |  of  1  sq,  rd.,  or  -f$  of  5  sq.  rd., 
and  cost  -fa  as  much  as  the  latter.  Add  the  cost  of  the  several 
aliquot  parts  to  the  cost  of  28  acres.  The  result  is  the  cost  of  the 
entire  tract  of  land. 

PROBLEM. — A  man  travels  3  mi.  20  rd.  5  yd.  in  1  hr.; 
how  far  will  he  go  in  6  da.  9  hr.  18  min.  45  sec.  (12  hr. 
to  a  day)? 

OPERATION. 

mi.      rd.      vd. 


OPERATION. 

$16 

28 

128 
32 

120     sq.  rd.  ==  f 
2  0     sq.  rd.  =  £ 
5     sq.  rd.  =  | 
1  5  i  sq.  yd.  =  TV 

$448 
1  2 
2 

.50 

.05 

$462.55 


SOLUTION.  —  This 
example  is  solved 
like  the  preceding, 
except  that  here 
the  multiplications 
and  divisions  are 
performed  on  a 
Compound  instead 
of  a  Simple  Number. 


20 


6  da.     =8X9  hr. 
1  5  min.  =  J-  of  1  " 
3     "=  £  of  1  5  min. 

45  sec.    =  J  of      3  min. 


27 

188 

1 

220 

225 

2J 

245 

H 

49 

i 

» 

1  2 

1  A 

249         80 


ALIQUOT  PARTS.  173 

NOTE.  —  In  all  questions  in  aliquot  parts,  one  of  the  numbers 
indicates  a  rate,  and  the  other  is  a  Compound  Number  whose  value 
at  this  rate  is  to  be  found. 

Rule  for  Aliquot  Parts.  —  Multiply  the  number  indicating 
tJie  rate  by  the  number  of  that  denomination  for  whose  unit  the 
rate  is  given,  and  separate  the  numbers  of  the  other  denomina- 
tions into  parts  whose  values  can  be  obtained  directly  by  a 
simple  division  or  multiplication  of  one  of  the  preceding  values. 
Add  these  different  values;  the  result  will  be  the  entire  value 
required. 

NOTES.  —  1.  Sometimes  one  of  the  values  may  be  obtained  by 
adding  or  subtracting  two  preceding  values  instead  of  by  multiplying 
or  dividing. 

2.  Aliquot  parts  are  generally  used  in  examples  involving  U.  S. 
money,  and  the  following  table  should  be  memorized  for  future  use. 


5    - 


10   =  - 


ALIQUOT  PARTS  OF  100. 


12    = 


20    = 


25  = 
331  = 
50  = 


REMARK. — The  following  multiples  of  aliquot  parts  of  100,  are 
often  used:  18}  =  •&,  37*=  f,  40  =  f,  60  —  f,  62^  =  f,  75  =  f, 
87} =f 

EXAMPLES  FOR  PRACTICE. 

1.  If  a  man  travel  2  mi.   105  rd.   6£  ft.  in  1  hour,  how 
far  can  he  travel  in  30  hr.  29  min.  52  sec.? 

71  mi.  12  rd.  4  ft.  ^  in. 

2.  An  old  record  says  that  694  A.  1    R.  22   P.  of  land, 
at  $11.52  per  acre,  brought  $8009.344;  what  is  the  error  in 
the  calculation  ?  $10  too  much. 

3.  At  $15.46  an  oz.,  what  will  be  the  cost  of  7  Ib.  8  oz. 
16  pwt.   11  gr.  of  gold?  $1435.04 


1 74  RA  Y>  S  HIGHER  ARITHMETIC. 

4.  What  is  the  cost  of  88  gal.  3  qt.  1  pt.  of  vinegar,  at 
37|ct.  a  gallon?  $33.33— 

5.  If  the  heart  should  beat  97920  times  in  each  day,  how 
many  times  would  it  beat  in  8  da.  5  hr.  25  min.  30  sec.? 

805494  times. 

6.  At  a  cost  of  $8190.50  per  mile  over  the  plain,  and  at  a 
rate  of  $84480  per  mile  of  tunnel,  what  is  the  cost  of  a  rail- 
way 17  mi.  150  rd.  plain,  and  70  rd.  tunnel? 

$161557.80  nearly. 

7.  What  is  the  value  of  20  T.  1  cwt.  13  Ib.  of  sugar,  at 
$3f  per  cwt.?  $1504.2375 

8.  If  £3  6s.  silver  weigh  1  ft.  troy,  how  much  will  17  ft. 
11  oz.  16  pwt.  9  gr.  be  worth?  £59  7s.  + 

9.  If  a  steam-ship  could  make  16  mi.  67f  rd.  in  1  hr., 
how  far  could  it  go  in  24  da.  22  hr.  56  min.  12  sec.  ? 

9709  mi.  29.09  rd. 


Topical  Outline. 
OPERATIONS  WITH  COMPOUND  NUMBERS. 

f  1.  Definition. 

1.  Reduction J  2.  Cases... /  L  From  Higher  to  Lower. 

1  3   Rules  5,  2«  From  Lower  to  Higher. 

2.  Addition /I-  Definition. 

12. 


Rule. 

3.  Subtraction 1 1-  Definition. 

I  2.  Rule. 

4.  Multiplication    /  1-  Definition. 

I  2.  Rule. 

fl.  Definition. 

5.  Division J  2.  Cases. 

I  3.  Rule. 


•Case  I.— Rule. 


f< 

6.  Longitude  and  Time.  1  Case  II.— Rule. 

(.  Table  of  Longitudes. 
f  1.  Definition. 

7.  Aliquot  Parts -J  2.  Rule. 

(3.  Table. 


XII.  EATIO. 

DEFINITIONS. 

228.  1.  Ratio  is  a  Latin  word,  signifying  relation  or  con- 
nection; in  Arithmetic,  it  is  the  measure  of  the  relation  of 
one  number  to  another  of  the  same  kind,  expressed  by  their 
quotient. 

2.  A  Ratio  is  found  by  dividing  the  first  number  by  the 
second  ;  as,  the  ratio  of  8  to  4  is  2.     The  ratio  is  abstract. 

3.  The  Sign  of  Ratio  is  the  colon  (:),  which  is  the  sign 
of  division,  with    the  horizontal   line  omitted ;  thus,   6  :   4 
signifies  the   ratio  of  6  to  4  —  | . 

4.  Each  number  is  called  a  term  of  the  ratio,  and  both 
together  a   couplet  or  ratio.     The  first  term  of  a  ratio  is 
the  antecedent,  which  means  going  before;   the  second  term 
is  the  consequent,  which  means  followii ig. 

5.  A  Simple  Ratio  is   a  single   ratio   consisting  of  two 
terms ;  as,  3  :  4  =  f . 

6.  A  Compound  Ratio  is  the  product   of  two  or  more 

(3*7 
simple   ratios:    as,    < 

I  5  :  8 

7.  The  Reciprocal  of  a  Ratio  is  1  divided  by  the  ratio, 
or  the  ratio  inverted  ;  thus,  the  reciprocal  of  2  :  3,  or  f ,  is 

12  3 

~  Z  —  2 '• 

8.  Inverse  Ratio  is  the  quotient  of  the   consequent  di- 
vided  by  the  antecedent ;  thus,    f   is  the   inverse    ratio  of 
4  to  5. 

9.  The  Value  of   the  Ratio   depends  upon  the  relative 
size  of  the  terms. 

(175) 


176  RAY'S  HIGHER  ARITHMETIC. 

229.  From  the  preceding  definitions  the  following  prin- 
ciples are  derived : 

Antecedent 

PRINCIPLES. — 1.  Ratio  =  -—-         —  • 

Consequent 

2.  Antecedent  =  Consequent  X  Ratio. 

Antecedent 

0.  Consequent  =  — — — ; 

Ratio 

Hence,  by  Art.  87 : 

1.  The  Ratio  is  multiplied  by  multiplying  the  Antecedent  or 
dividing  the  Consequent. 

2.  The  Ratio  is  divided  by  dividing  the  Antecedent  or  mul- 
tiplying the  Consequent. 

3.  The   Ratio    is    not    changed    by  multiplying    or   dividing 
both  terms  by  the  same  number. 

General  I&w*—Any  change  in  the  Antecedent  produces  a 
like  change  in  the  Ratio,  but  any  change  in  the  Consequent 
produces  an  opposite  change  in  the  Ratio. 

PROBLEM. — What  is  the  ratio  of  15  to  36  ? 

OPERATION. 

15  :  36  =  H  =  A- 
Rule. — Divide  the  Antecedent  by  the  Consequent. 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  ratio  of  2  ft.  6  in.  to  3  yd.  1  ft.  10  in.?      T%. 

2.  What    is    the    ratio    of    4    mi.    260    rd.     to     1    mi. 
96  rd.?  fff. 

3.  What  is  the  ratio   of  13  A.  145  sq.  rd.  :  6  A.  90  sq. 
rd.?  f|, 

4.  What  is  the  ratio  of  3  ft.  10  oz.  6  pwt.  KH-  gr.  :  2  ft. 
14f  pwt.? 


PROPORTION. 

5.  What  is  the   ratio  of    10  gal.    1.54  pt.  :  7  gal.  2  qt. 
.98  pt.?  £$|f 

6.  What  is  the  ratio  of  56  bu.  2  pk.  1  qt.  :  35  bu.  3  pk. 
6.055  qt.?  flM- 

7.  If  the   antecedent   is  7  and  the  ratio  1^-,  what  is  the 
consequent  ?  4f . 

8.  If  the  consequent  is   f  and  the   ratio  f ,  what   is   the 
antecedent  ?  ^2r. 

9.  What  is  the  ratio  of  a  yard  to  a  meter,  and  of  a  meter 
to  a  yard? 

10.  What  is  the  ratio  of  a  pound  avoirdupois  to  a  pound 
troy?  Uf- 

11.  Find    the    difference    between    the    compound    ratios 

{\\\}«*  fill-  H- 

12.  Find  the  difference  between  the  ratio  4|  :  7-|  and  the 
inverse  ratio.  yf  M* 

13.  If  the  consequent  is  6^,  and  the  ratio  is  2^,  what  is 
the  antecedent,  and  what    is  the  inverse    ratio  of  the   two 
numbers?  Antecedent  14f,  inverse  ratio  ^. 


XIII.   PROPORTION. 

DEFINITIONS. 

230.     1.  Proportion  is  an  equality  of  ratios. 

Thus,  4  :  6  :  :  8  :  12  is  a  proportion,  and  is  read  4  is  to  6  as  8 
is  to  12. 

2.  The  Sign  of  Proportion  is  the  double  colon  (  :  :  ). 

NOTE. — It  is  the  same  in  effect  as  the  sign  of  equality,  which  is 
sometimes  used  in  its  place. 

3.  The   two   ratios   compared  are  called   couplets.     The 
first  couplet  is  composed  of  the  first  and  second  terms,  and 
the  second  couplet  of  the  third  and  fourth  terms. 


178  RAY'S  HIGHER  ARITHMETIC. 

4.  Since  each   ratio   has  an   antecedent    and  consequent, 
every  proportion   has  two  antecedents  and  two  consequents, 
the  1st  and  3d  terms  being  the  antecedents,  and  the  2d  and 
4th  the  consequents. 

5.  The  first  and  last  terms  of  a  proportion  are  called  the 
extremes ;  the  middle  terms,  the  means.     All  the   terms 
are  called  proportionals,  and  the  last  term  is  said  to  be  a 
fourth  proportional  to  the  other  three  in  their  order. 

6.  When  three  numbers  are  proportional,  the  second  num- 
ber is  a  mean  proportional  between  the  other  two. 

Thus,  4:6  :  :  6:9;  six  is  a  mean  proportional  between  4  and  9. 

7.  Proportion  is   either   Simple   or  Compound :    Simple 
when  both  ratios  are  simple ;  Compound  when   one  or  both 
ratios  are  compound. 

PRINCIPLES. — 1.  In    every    proportion   the    product   of  the 
means  is  equal  to  the  product  of  the  extremes. 

2.  The  product  of  the  extremes   divided  by  either  mean,  will 
give  the  other  mean. 

3.  The  product  of  the  means  divided  by  either  extreme,  will 
give  the  other  extreme. 


SIMPLE  PEOPOKTIOK 

231.     1.  Simple  Proportion  is  an  expression  of  equality 
between  two  simple  ratios. 

2.  It  is  employed  when  three  terms  are  given  and  we  wish 
to  find  the  fourth.     Two  of  the  three  terms   are  alike,  and 
the  other  is  of  the  same  kind  as  the  fourth  which  is  to  be 
found. 

3.  All  proportions  must  be  true  according  to  Principle  1, 
which  is  the  test.     Principles  2  and  .3  indicate  methods  of 
finding  the  wanting  term. 


SIMPLE  PROPORTION.  179 

4.  The    Statement    is    the    proper    arrangement    of   the 
terms  of  the  proportion. 

PROBLEM. — If  6   horses  cost  $300,    what  will   15  horses 
cost? 

STATEMENT. 

6  horses  :  15  horses  ::  $300  :  ($     ). 

OPERATION. 

$300X15  „  _ 

6 

•  SOLUTION. — Since  6  horses  and  15  horses  maybe  compared,  they 
form  the  first  couplet;  also,  $300  and  $  —  may  be  compared,  as  they 
are  of  the  same  unit  of  value. 

NOTES. — 1.  To  find  the  missing  extreme,  we  use  Prin.  3. 
2.  To  prove  the  proportion,  we  use  Prin.  1.    Thus,  $750  X  6  "= 
$300  X  15. 

PROBLEM. — If  15  men  do  a  piece  of  work  in  9f  da.,  how 
long  will  36  men  be  in  doing  the  same? 

STATEMENT. 

SOLUTION— Since  36  men  will  men    men     da      da> 

require  less  time  than  15  men  to  36*15-'94-() 
do    the    same   work,   the    answer 

should  be  less  than  9-f  da.;  make  a  OPERATION. 

decreasing  ratio,  Jf,  and  multiply  9f  =-V-  da. 

the  remaining  quantity  by  it.  -^  X  if  ==  4  da.,  Ans. 

Rule. — 1.   For  the  third  term,   write  that    number  which  is 
of  the  same  denomination  as  the  number  required. 

2.  For  the   second    term,  write  the  GREATER  of  the  two  re- 
maining numbers,  when  the  fourth  term  is   to  be    greater  than 
the  third;    and    the   LESS,    when   the  fourth    term  is  to  be  less 
than  the  third. 

3.  Divide  the  product   of  the  second  and  third  terms   by  Hie 
first;  the  quotient  will  be  the  fourth  term,  or  number  required. 


1 80  RA  Y>  S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

NOTE. — Problems    marked   with    an   asterisk   are   to    be  solved 
mentally. 

1.*  If  I  walk  101  mi.  in  3  hr.,  how  far  will  I  go  in  10 
hr.,  at  the  same  rate?  35  mi. 

2.  If  the  fore-wheel   of  a  carriage  is  8   ft.   2   in.   in  cir- 
cumference, and  turns  round  670  times,  how  often  will  the 
hind-wheel,  which   is   11    ft.    8    in.    in  circumference,   turn 
round  in  going  the  same  distance?  469  times. 

3.  If  a  horse  trot  3  mi.  in  8  min.  15  sec.,  how  far  can 
he  trot  in  an  hour,  at  the  same  rate  ?  21T9T  mi. 

4.  What  is  a  servant's  wages  for  3  wk.   5  da.,  at  $1.75 
per  week?  $6.50 

5.  What  should  be  paid  for  a  barrel  of  powder,  containing 
132  lb.,  if  15  Ib.  are  sold  for  $5.43|?  $47.85 

6.  A  body  of  soldiers  are  42  in  rank  when  they  are  24 
in  file :  if  they  were  36    in   rank,  how  many  in  file  would 
there  be?  28. 

7.  If  a  pulse  beats  28  times  in  16  sec.,  how  many  times 
does  it  beat  in  a  minute?  105  times. 

8.  If  a  cane  3  ft.  4  in.  long,  held  upright,  casts  a  shadow 
2  ft.  1  in.  long,  how  high   is  a  tree  whose    shadow  at    the 
same  time  is  25  ft.  9  in.  ?  41  ft.  2f  in. 

9.  If  a  farm  of  160  A.  rents  for  $450,  how  much  should 
be  charged  for  one  of  840  A  ?  $2362.50. 

10.  A  grocer  has  a  false  gallon,  containing  3  qt.  1|-  pt.: 
what  is  the  worth  of  the  liquor  that  he  sells  for  $240,  and 
what  is  his  gain  by  the  cheat?  $225,  and  $15  gain. 

11.  If  he  uses  14f  oz.   for   a  pound,  how  much  does  he 
cheat  by  selling  sugar  for  $27.52?  $2.15 

12.  An  equatorial  degree  is  365000  ft.:  how  many  ft.  in 
80°  24'  37"  of  the  same?  29349751T?8-  ft. 

13.  If  a    pendulum    beats   5000  times  a  day,  how  often 
does  it  beat  in  2  hr.  20  min.   5  sec.  ?  486^-|f  times. 


SIMPLE  PROPORTION.  181 

14.*  If  it  takes  108  days,  of  8|-  hr.,  to  do  a  piece  of  work, 
how  many  days  of  6f  hr.  would  it  take  ?  136  days. 

15.  A  man  borrows  $1750,  and  keeps  it  1  yr.  8  mon.: 
how  long  should  he  lend  $1200  to  compensate  for  the 
favor  ?  2  yr.  5  mon.  5  da. 

l(j.  A  garrison  has  food  to  last  9  mon.,  giving  each  man 
1  Ib.  2  oz.  a  day :  what  should  be  a  man's  daily  allowance, 
to  make  the  same  food  last  1  yr.  8  mon.?  8T^  oz. 

17.  A  garrison  of  560  men  have  provisions  to  last  during 
a  siege,  at  the  rate  of  1  Ib.    4  oz.  a  day  per  man ;  if  the 
daily  allowance  is  reduced  to  14  oz/  per  man,  how  large  a 
reinforcement  could  be  received?  240  men. 

18.  A  shadow  of  a  cloud  moves  400  ft.  in  18f  sec.:  what 
was  the  wind's  velocity  per  hour?  14T6T  mi. 

19.  If  1  Ib.  troy  of  English  standard  silver  is  worth  £3 
6s.,  what  is  1  Ib.  av.  worth?  £4  2|d. 

20.  If  I  go  a  journey  in  12f  days,  at  40  mi.  a  day,  how 
long  would  it  take  me  at  29f  mi.   a  day  ?  17^  da. 

21.*  If  |  of  a  ship  is  worth  $6000,  what  is  the  whole 
of  it  worth  ?  $10800. 

22.  If  A,  worth  $5840,  is  taxed  $78.14,  what  is  B  worth, 
who  is  taxed  $256.01?  $19133.59— 

23.*  What  are  4  Ib.  6  oz.  of  butter  worth,  at  28  ct.  a 
Ib.?  $1.221 

24.  If  I   gain   $160.29  in  2  yr.  3  mon.,   what    would  I 
gain  in  5  yr.  6  mon.,  at  that  rate?  $391.82 

25.  If  I  gain  $92.54  on  $1156.75  worth  of  sugar,  how 
much  must  I  sell  to  gain  $67.32?  $841.50  worth. 

26.  If  coffee   costing  $255   is   now  worth  $318.75,  what 
did  $1285.20  worth  cost?  $1028.16 

27.  A  has  cloth  at  $3.25  a  yd.,  and  B  has  flour  at  $5.50 
a  barrel.     If,  in  trading,  A  puts  his  cloth  at  $3.62^,  what 
should  B  charge  for  his  flour?  $6.13T63 

28.*  If  a  boat  is  rowed  at  the  rate  of  6  miles  an  hour, 
and  is  driven  44  feet  in  9  strokes  of  the  oar,  how  many 
strokes  are  made  in  a  minute?  108  strokes. 


182  RAY'S  HIGHER  ARITHMETIC. 

29.  If  I   gain    $7.75  by  trading   with    $100,  how  much 
ought  I  to  gain  on  $847.56?  $65.6859 

30.  What    is  a  pile  of   wood,  15    ft.    long,   10£  ft.  high, 
and  12ft.  wide,  worth,  at  $4.25  a  cord?  $62.75 

REMARK. — In  Fahrenheit's  thermometer,  the  freezing  point  of 
water  is  marked  32°,  and  the  boiling  point  212° :  in  the  Centigrade, 
the  freezing  point  is  0°,  and  the  boiling  point  100°:  in  Reaumer's, 
the  freezing  point  is  0°,  and  the  boiling  point  80°. 

31.  From  the  above  data,  find  the  value  of  a  degree  of 
each  thermometer  in  the  degrees  of  the  other  two. 

1°  F.  =±°  R=f°  C.;  1°  C.  =lf°  P.= 
-|°  R;  1°  R.  =1J°  C.  =  2i°  F. 

32.  Convert  108°  F.   to  degrees  of  the  other   two  ther- 
mometers. 33£°  R.  and  42f  °  C. 

33.  Convert  25°  R.  to  degrees  of  the  other  two  thermom- 
eters. 31^°  C.  and  88^°  F. 

34.  Convert   46°    C.    to   degrees   of   the  other  two  ther- 
mometers. 36f°  R.  and  114f°  F. 

REMARKS. — 1.  In  the  working  of  machinery,  it  is  ascertained  that 
the  available  power  is  to  the  weight  overcome,  inversely  as  the  distances  they 
pass  over  in  the  same  time. 

2.  Inverse  variation  exists    between   two   numbers   when   one   in- 
creases as  the  other  decreases. 

3.  The  available  power  is  taken   f  of  the  whole  power,  |-  being 
allowed  for  friction  and  other  impediments. 

'  35.  If  the  whole  power  applied  is  180  Ib.  and  moves  4  ft., 
how  far  will  it  lift  a  weight  of  960  Ib.  ?  6  in. 

36.  If  512  Ib.  be  lifted   1  ft.   3  in.  by  a  power  moving 
6  ft.  8  in.,  what  is  the  power  ?  144  Ib. 

37.  A  lifts  a  weight  of  1410  Ib.  by  a  wheel  and  axle;  for 
every  3  ft.  of  rope  that  passes  through  his  hands  the  weight 
rises  4|-  in.:  what  power  does  he  exert?  270  Ib. 

38.  A  man  weighing  198  Ib.  lets  himself  down  54  ft.  with 
a  uniform  motion,  by  a  wheel  and  axle :  if  the  weight  at  the 
hook  rises  12  ft.,  how  much  is  it?  594  Ib. 


SIMPLE  PROPORTION.  183 

39.  Two   bodies   free   to   move,  attract    each    other    with 
forces  that  vary  inversely  as  their  weights.     If  the  weights 
are  9  Ib.  and  4  lb.,  and  the  smaller  is  attracted  10  ft.,  how 
far  will  the  larger  be  attracted?  4  ft.  5^  in. 

40.  Suppose  the  earth  and  moon  to  approach  each  other 
in  obedience  to  this  law,  their  weights  being  49147  and  123 
respectively,  how  many  miles  would  the  moon   move  while 
the  earth  moved  250  miles  ?  99892+  mi. 

Can  the  three  following  questions  be  solved  by  pro- 
portion? 

41.  If  3   men    mow  5  A.  of  grass  in  a   day,   how  many 
men  will  mow  13^  A.  in  a  day? 

42.*  If  6  men  build  a  wall  in  7  da.,  how  long  would  10 
men  be  in  doing  the  same? 

43.*  If  I  gain  15  cents  each,  by  selling  books  at  $4.80  a 
doz.,  what  is  my  gain  on  each  at  $5.40  a  doz.? 

44.  A  clock  which   loses  5  minutes  a  day,  was  set  right 
at  6  in  the  morning  of  January  1st :  what  will  be  the  right 
time  when  that  clock  points  to  11  on  the  15th? 

11  min.  17.35+  sec.  past  noon. 

45.  If  water  begin  and  continue  running  at  the  rate  of 
80  gal.  an  hour,  into  a  cellar  12  ft.  long,  8  ft.  wide,  and  6 
ft.  deep,  while  it  soaks  away  at  the  rate  of  35  gal.  an  hour, 
in  what  time  will  the  cellar  be  full?  95.75+  hr. 

46.  Take  the  proportion  of  4  :  9  : :  252  :  a  fourth  term. 
If  the  third  and  fourth  terms  each  be  increased  by  7,  while 
the  first  remains  unchanged,  what  multiplier  is  needed  by 
the  second  to  make  a  proportion?  |||. 

47.  Prove  that  there  is  no  number  which  can  be  added 
to  each  term  of  6  :  3  : :   18  :  9  so  that  the  resulting  num- 
bers shall  stand  in  proportion. 

48.  A  certain  number  has  been  divided  by  one  more  than 
itself,  giving  a  quotient  \\  what  is  the  number?  J. 

49.  If  48  lb.    of  sea-water  contain  1^-  lb.   of  salt,  how 
much  fresh  water  must  be  added  to  these  48  lb.  so  that  40 
lb.  of  the  mixture  shall  contain  \  lb.  of  salt?  72  lb. 


184  RA  Y'S  HIGHER  ARITHMETIC. 


COMPOUND  PKOPOKTION. 

232.  Compound  Proportion  is  an  expression  of  equality 
between  two  ratios  when  either  or  when  each  ratio  is  Com- 
pound. 

PROBLEM. — If  3  men  mow  8   A.   of  grass  in  4  da.,  how 
long  would  10  men  be  in  mowing  36  A.? 

STATEMENT. 

1  0  men  :  3    men  ) 

SOLUTION.— Since  Q    A  Q  a    \     I  :  :   4  da'   :    (  )  da' 

o   xx.        .    O  D    A.*  ) 
the       denomination 

of  the  required  term  OPERATION. 

is   days,    make    the  9 

third  term  4  da.     In  ^  jj 

forming  the  first  and  3  Xj!  0  X  jft 

second    terms,    con-  1  0  v  $ =   *    =     ^        '        S' 

sider    each   denom-  ^  y/  A  p 

ination    separately ; 

10  men  can  do  the  same  amount  of  work  in  less  time  than  3  men ; 
hence,  the  first  ratio  is,  10  men  :  3  men,  the  less  number  being  the 
second  term.  Since  it  takes  4  da.  to  mow  8  A.,  it  will  take  a  greater 
number  of  days  to  mow  36  A.,  and  the  second  ratio  is,  8  A.  :  36  A., 
the  greater  number  being  the  second  term.  Then  dividing  the  con- 
tinued product  of  the  means  by  that  of  the  extremes  (Art.  230, 
Prin.  3),  after  cancellation,  we  have  5f  da.,  the  required  term. 

Rule. — 1.  For  the  third  term,    write  that   number  which  is 
of  the  same  denomination  as  the  number  required. 

2.  Arrange  each  pair   of  numbers   having  the   same    denom- 
ination in  the  compound  ratio,  as  if,  with  the  third  term,  they 
formed  a  simple  proportion. 

3.  Divide  the  product  of  the  numbers  in  the  second  and  third 
term,s  by  the  product  of  the  numbers  in  the  first  term:  the  quotient 
will  be  the  required  term. 

233.  Problems    in    Compound    Proportion    are    readily 
solved   by  separating  all   the  quantities  involved    into    two 
causes  and  two  effects. 


COMPOUND  PROPORTION.  185 

PROBLEM.  —If  6  men,  in  10  days  of  9  hr.  each,  build 
25  rd.  of  fence,  how  many  hours  a  day  must  8  men  work 
to  build  48  rd.  in  12  days? 

SOLUTION.  —  6  men  10  da.  and  9  hr.  constitute  the  first  cause, 
whose  effect  is  25  rd.  ;  8  men  12  da.  and  (  )  hr0  constitute  the  second 
cause,  whose  effect  is  48  rd.  Hence, 

STATEMENT. 

8  men.  "| 

12  da.  V::  25  rd.  :  48  rd. 

(  )  hr.J 

OPERATION. 


Rule  of   Cause  and  Effect.  —  1.    Separate  all   the  quan- 
tities contained  in  the  question  into  two  causes  and  their  effects. 

2.  Write,  for  the  first  term  of  a  proportion,  all  the  quan- 
tities that  constitute  the  first  cause  ;  for  the  second  term,  all  that 
constitute  the  second  cause;  for  the  third,  all  that  constitute  the 
effect  of  the  first  cause  ;  and  for  the  fourth,  all  that  constitute  the 
effect  of  the  second  cause. 

3.  The   required   quantity   may    be    indicated   by  a    bracket, 
and  found  by  Art.  230,  Principles. 

NOTE.  —  The  two  causes  must  be  exactly  alike  in  the  number  and 
kind  of  their  terms  ;  and  so  must  the  two  effects. 


EXAMPLES  FOR  PRACTICE. 

1.  If  18  pipes,  each  delivering  6  gal.  per  minute,  fill  a 
cistern  in  2  hr.  16  min.,  how  many  pipes,  each  delivering 
20  gal.  per  minute,  will  fill  a  cistern  7^  times  as  large  as 
the  first,  in  3  hr.  24  min.  ?  27  pipes. 

H.  A.  16. 


186  RA  Y'S  HIGHER  ARITHMETIC. 

2.  The  use  of  $100  for  1  year  is  worth  $8  :  what  is  the 
use  of  $4500  for  2  yr.  8  mon.  worth  ?  $960. 

3.  If  12  men  mow  25  A.  of  grass  in  2  da.  of  10^  hr., 
how  many  hours  a  day  must  14  men  work  to  mow  an  80  A. 
field  in  6  days  ?  9f  hr. 

4.  If  4  horses  draw  a  railroad  car  9  miles  an  hour,  how 
many  miles  an  hour  can  a  steam  engine  of  150  horse-power 
drive   a  train   of  12  such  cars,   the  locomotive  and  tender 
being  counted  3  cars  ?  22^  mi.  per  hr. 

5.  If  12   men,  working   20  days   10   hours   a  day,  mow 
247.114    hektars  of  timothy,    how   many  men   in   30  days, 
working    8   hours   a  day,    will    mow    1976912    centars    of 
timothy  of  the  same  quality  ?  8  men. 

6.  If  the   use   of   $3750   for    8    mon.   is  worth   $68.75, 
what    sum  is   that   whose   use   for    2    yr.    4   mon.  is  worth 
$250?  $3896.10+ 

7.  If  the  use  of  $1500  for  3  yr.  8  mon.  25  da.  is  worth 
$336.25,  what  is  the  use  of  $100  for  1  yr.  worth?  $6. 

8.  A  garrison    of  1800   men   has    provisions  to  last    4|- 
months,  at  the  rate  of  1  Ib.  4  oz.  a  day  to  each :  how  long 
will  5  times  as  much  last  3500  men,  at  the  rate  of  12  oz. 
per  day  to  each  man?  1  yr.  7^  months. 

9.  What   sum   of  money  is  that  whose  use  for  3   yr.,  at 
the  rate  of  $4-|  for  every  hundred,  is  worth  as  much  as  the 
use  of  $540  for  1  yr.  8  mon.,  at  the  rate  of  $7  for  every 
hundred?  $466. 66| 

10.  A  man  has  a  bin  7  ft.  long  by  2^  ft.  wide,  and  2  ft. 
deep,  which  contains  28   bu.  of  corn  :  how    deep   must   he 
make  another,  which  is  to  be  18  ft.  long  by  1|-  ft.  wide,  in 
order  to  contain  120  bu.  ?  4|  ft. 

11.  If  it  require  4500  bricks,  8  in.  long  by  4  in  wide,  to 
pave  9.  court-yard   40  ft.    long  by  25   ft.   wide,   how  many 
tiles,  10  in.   square,  will   be    needed  to  pave  a  hall  75  ft. 
long  by  16  ft.  wide?  1728  tiles. 

12.  If   150000   bricks  are  used  for   a  house  whose  walls 
average  1^  ft.  thick,  30   ft.    high,  and   216  ft.   long,    how 


COMPOUND  PROPORTION.  187 

many  will  build  one  with  walls  2  ft.  thick,  24  ft.  high,  and 
324ft.  long?  240000  bricks. 

13.  If  240  panes  of  glass  18  in.  long,  10  in.  wide,  glaze  a 
house,    how   many  panes   16  in.   long    by   12  in.  wide  will 
glaze  a  row  of  6  such  houses?  1350  panes. 

14.  If  it  require  800  reams  of  paper  to  publish  5000  volumes 
of  a  duodecimo  book  containing  320  pages,  how  many  reams 
will  be   needed  to   publish  24000  copies    of  a  book,  octavo 
size,  of  550  pages  ?  9900  reams. 

15.  If   15   men  cut  480  sters  of  wood   in  10  days,  of  8 
hours  each:  how  many  boys  will  it   take  to  cut  1152  sters 
of  wood,  only  f  as  hard,  in  16  days,  of  6  hours  each,  pro- 
vided that  while  working  a  boy  can  do  only  f  as  much  as 
a  man,  and  that  ^  of  the  boys  are  idle  at  a  time  throughout 
the  work  ?  24  boys. 


Topical   Outline. 
RATIO  AND  PROPORTION. 


II.  Definitions. 
2.  Principles. 
3.  General  Law. 

|    1.  Definitions. 
2.  Principles. 
2.  Proportion...  \   3.  Kinds /  1-  Simple. 

4.  Statement.        *  2-  Compound. 

5.  Rules. 


XIY.    PEKCENTAGE. 

DEFINITIONS. 

234.  1.  Percentage   is  a   term    applied   to   all   calcula- 
tions  in   which    100  is  the  basis  of  comparison;    it  is  also 
used   to  denote    the   result    arising    from   taking    so    many 
hundredths  of  a  given  number. 

2.  Per    Cent    is    derived    from    the    Latin    phrase   per 
centum,  which  means  by  or  on  the  hundred. 

3.  The  Sign  of  Per  Cent  is    <fa\    the  expression  4%  —.. 
T|~o,  is  read  "4  per  cent"  equals  y^;  or,  decimally,  .04 

4.  The  elements  in  Percentage   are   the    Base,  the  Bate, 
the  Percentage,  and  the  Amount  or  Difference. 

5.  The  Base  is  the   number  on  which  the  Percentage  is 
estimated. 

6.  The  Rate  is  the   number  of  hundredths  to  be  taken. 

7.  The   Percentage    is    the   result    arising    from    taking 
that  part  of  the  Base  expressed  by  the  Rate. 

8.  The  Amount  is  the  Base  plus  the  Percentage. 

9.  The  Difference  is  the  Base  minus  the  Percentage. 

NOTATION. 

235.  It  is  convenient  to  use  the  following  notation: 

1.  Base  =B. 

2.  Rate  =  R. 

3.  Percentage  —  P. 
4      f  Amount  =  A. 

I  Difference  =  D. 

(188.) 


PEE  CENT  A  OK  189 

REMARK. — All  problems  in  Percentage  refer  to  two  or  more  of 
the  above  terms.  Owing  to  the  relations  existing  among  these 
terms,  any  two  of  them  being  given  the  others  can  be  found.  These 
relations  give  rise  to  the  following  cases  : 


CASE    I. 

236.     Given  the  base  and  the  rate,  to  find  the  per- 
centage. 

PRINCIPLE.  —  The  percentage   of  any   number   is  the   same 
part  of  that  number  as  the  given  rate  is  of  100^. 

PROBLEM. — If  I  have  160  sheep,  and  sell  35%  of  them: 
how  many  do  I  sell? 

OPERATIONS. 

1.  160  sheep  X-3  5  —  56  sheep,  Ans. 

^ 

2.  10 0^  =  160  sheep. 

1^  =  1.60  sheep. 
tf0  =  1.60  X  35  =  56  sheep,  Ans. 


3.     1  6  0  X  A  =  5  6  sheep,  Ans.  (Art.  218,  Kern.) 
SUGGESTION.— 3  5^  =  f-fc  =  &. 

SOLUTION. — Take  .35  of  the  base  ;  the  result  is  the  percentage. 
FORMULA.— B  X  R  =  P. 

Rule  1. — Multiply   the  base  by  the  rate   ft    expressed   deci- 
mally;  the  product  is  the  percentage. 

Rule  2. — Find  that  part  of  the  base  which  the  rate  %   is 
of  100. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  62^  of  1664  men.  1040  men. 

2.  Find  35%  off  TV 

3.  Find  9f  %  of  48  mi.  256  rd.  4  mi.  184  rd. 


190  RAY'S  HIGHER  ARITHMETIC. 

4.  Find  111%  of  $3283.47  $364.83 

5.  Find  33|$  of  127  gal.  3  qt.  1  pt.       42  gal.  2  qt.  1  pt 

6.  Find  98%   of  14  cwt.  2  qr.  20  Ib. 

14  cwt.  1  qr.   15f  Ib. 

7.  Find  40%   of  6  hr.  28  min.   15  sec. 

2  hr.  35  min.  18  sec. 

8.  Find  104%  of  75  A.  75  sq.  rd.         78  A.  78  sq.  rd. 

9.  Find  15f  %  of  a  book  of  576  pages.  90  pages. 

10.  Find  56^%  Of  144  cattle.  81  cattle. 

11.  Find  16f%  of  1932  hogs.  322  hogs. 

12.  Find  1000%  of  $5.43f  $54.371 

13.  Find  21%  of  i.  sV 

14.  What  part  is  25%  of  a  farm?  1. 

15.  What    part   of  a    quantity  is  18f%   of  it;     31-1%; 
37i%;  43f%;  561%;  621%;  68f%;  811%;  83£%;  871%; 

93f  %  ?  A,  T5c >  I,  T7e >  T96 >  f  >  «,  it,  I,  *,  if  of  it. 

16.  How   much   is    100%    of   a   quantity;    125%   of   it; 
250%;  675%;  1000^;  9437^? 

1  time,  1^,2^,  6f,  10,  94|  times  the  quantity. 

17.*  A  man  owning  f  of  a  ship,  sold  40^  of  his  share : 
what  part  of  the  ship  did  he  sell,  and  what  part  did  he 
still  own?  -^  sold  ;  -fa  left. 

18.*  A  owed  B  a  sum  of  money ;  at  one  time  he  paid 
him  40%  of  it;  afterward  he  paid  him  25%  of  what  he 
owed ;  and  finally  he  paid  him  20^  of  what  he  then  owed : 
how  much  does  he  still  owe?  -f^  of  it. 

19.  Out  of  a  cask  containing  47  gal.  2  qt.  1  pt.,  leaked 
6f %:  how  much  was  that?  3  gal.  If  pt. 

20.  A  has  an  income  of  $1200  a  year;  he  pays  23%  of 
it  for  board;   10f^  for  clothing;  6f^   for  books;  ^%  for 
newspapers;   12|-%   for  other  expenses:   how  much  does  he 
pay  for  each  item,  and  how  much  does  he  save  at  the  end 
of  the  year?  $276,  bd.;  $124.80,  cl.;  $81,  bks.;  $7,  npr. 

$154.50,  other  ex.;  $556.70  saved. 

21.  Find  10%  of  20%  of  $13.50  27  ct. 

22.  Find  40%  of  15%  of  75%  of  $133.331  $6. 


PERCENTAGE.  191 

23.  A  man  contracts  to  supply  dressed  stone  for  a  court- 
house for  $119449,  if  the  rough  stone  costs  him.-  16   ct.    a 
cu.    ft.;  but  if  he  can  get  it  for  15  ct.    a   cu.   ft.,  he  will 
deduct    3ft>   from   his  bill ;    how   many    cu.    ft.    would    be 
needed,  and  what  does  he  charge  for  dressing  a  cu.  ft.  ? 

358347  cu.  ft.,  and  17£  ct.  a  cu.  ft. 

24.  48%  of  brandy  is  alcohol;  how  much  alcohol  does  a 
man  swallow  in  40  years,  if  he  drinks  a  gill  of  brandy  3 
times  a  day?  657  gal.  1  qt.  1  pt.  2.4  gills. 

25.  A   had  $1200;  he  gave   30%   to  a  son,  20%  of  the 
remainder  to  his  daughter,  and   so  divided   the  rest  among 
four  brothers  that  each  after  the  first  had  $12  less  than  the 
preceding:  how  much  did  the  last  receive?  $150. 

26.  What  number  increased  by  20^  of  3.5,   diminished 
by  12|%  of  9.6,  gives  3£?  4. 

CASE    II. 

237.     Given   the   base   and  the    percentage,  to    find 
the  rate. 

PRINCIPLE. — The  rate  equals  the  number  of  hundredths  that 
the  percentage  is  of  the  base. 

PROBLEM. — What  per  cent  of  45  is  9? 

SOLUTION. — 9  is  J  of  45  ;  but  i  of  any  OPERATION. 

number  is  equal  to  20^  of  that  number;  ^  =  1^20^,  Ans. 

hence,  9  is  20  ft  of  45. 

p 

FORMULA. —  —  •=  R. 

JL> 

Rule  1. — Divide  the  percentage  by  the  base;  the  quotient  is 
the  rate ; 

Rule  2. — Find  that  part   of  100^  that  the    percentage  is 
of  the  base. 


192  RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  15  ct.  is  how  many  %  of  $2?  7$%. 

2.  2  yd.  2  ft.  3  in.  is  how  many  %  of  4  rd.? 

3.  3  gal.  3  qt.  is  what  %  of  31$  gal.? 

4.  |  is  how  many  %  of  f  ?  831%. 

5.  $  of  |  of  £  is  what  %  of  1TV? 


6.  is  how  many  £~&*?  222f  %. 

O  J-V/ 

7.  $5.12  is  what  %  of  $640?  f%. 

8.  $3.20  is  what  %  of  $2000? 

9.  750  men  is  what  %  of  12000  men? 

10.  3  qt.  11  pt.  is  what  %  of  5  gal.  2-*-  qt.?  16f%. 

11.  A's  money  is  50%  more  than  B's;  then  B's  money  is 
how  many  %  less  than  A's  ? 

12.  What  %  of  a  number  is  8%  of  35%  of  it? 

13.  What  %  of  a  number  is  2$%  of  2^%  of  it?  %. 

14.  What  %  of  a  number  is  40^  of  Q2$%  of  it?      25%. 

15.  12%  of  $75  is  what  %  of  $108?  $$%. 

16.  U.  S.  standard  gold  and  silver  are  9  parts  pure  to  1 
part  alloy:  what  %  of  alloy  is  that? 

17.  What  %  of  a  meter  is  a  yard? 

18.  How  many  ^   of  a  township  6  miles  square,  does  a 
man  own  who  has  9000  acres  ?  ^TQ%  • 

19.  How  many  %   of  a  quantity  is  40^  of  25%    of  it  ? 
also,  16%  of  yi\%  of  it?  also,  4£%  of  120%  of  it?  also, 
2%  of  80^  of  66f  %  of  it?  also,  f  %  of  36%  of  75%  of  it? 
also,  6|%  of  221%  Of  96%  of  it? 

10,  6,  5,  1TV,  5Vo>  Wo^- 

20.  30%   of  the  whole  of  an  article  is  how  many  %  of  f 
of  it?  45%. 

21.  25%  of  |  of  an  article  is  how  many  %  of  f  of  it? 

13i%. 

22.  How  many  %  of  his  time  does  a  man  rest,  who  sleeps 
7  hr.  out  of  every  24  ?  291%  . 


PERCENTAGE.  193 


CASE   III. 

238.     Given   the    rate   and   the   percentage,  to   find 
the  base. 

PRINCIPLE.  —  The  base  bears  the  same  ratio  to  the  percentage 
that  100%  does  to  the  rate. 

PROBLEM.  —  95  is  5%  of  what  number? 

OPERATION. 

5#=  A  =95; 

i*«?tfc=i*5 

100  ^  =  ^  =  19X100  =  1900,  .4ns. 
Or,  9  5  -J-  .  0  5  =  1  9  0  0,  Ana. 

p 
FORMULA.  —  —  =  B. 


Rule  1.  —  Divide   the  percentage  by  the  rate,  and  then  mul- 
tiply the  quotient  by  100  ;  the  product  is  the  base. 

Rule  2.  —  Divide  the  percentage  by  the   rate   expressed  deci- 
mally ;  the  quotient  is  the  base. 


EXAMPLES  FOR  PRACTICE. 

1.  $3.80  is  5^  of  what  sum?  $76. 

2.  T2T  is  80^  of  what  number  ?  -fa. 

3.  16  is  \\%   of  what  number?  1066f. 

4.  31£  ct.  is  15f^  of  what?  $2. 

5.  $10.75  is  8|$  of  what?  $322.50 

6.  162  men  is  4f  %  of  how  many  men?  3375  men. 

7.  $19.20  is  T6^  of  what?  $32000 

8.  $189.80  is  104%  of  what?  $182.50 

9.  16  gal.  1  pt.  is  §\%  of  what?  262  gal.  2  qt. 


10.  10  mi.  316  rd.  is  75^  of  what?        ,      14  mi.  208  rd. 

H.  A.  17. 


194  RAY'S  HIGHER  ARITHMETIC. 

11.  Thirty-six  men  of  a  ship's  crew  die,  which  is  421 L% 
of  the  whole :  what  was  her  crew  ?  84  men. 

12.  A  stock-farmer  sells  144  sheep,  which  is  12^%  of  his 
flock:  how  many  sheep  had  he?  1125  sheep. 

13.  A  merchant  sells  35%  of  his  stock  for  $6000:  what 
is  it  all  worth  at  that  rate?  $17142.86 

14.  I  shot  12  pigeons,  which  was  2f  %  of  the  flock:  how 
many  pigeons  escaped  ?  438  pigeons. 

15.  A,  owing  B,  hands  him  a  $10  bill,  and  says,  "  there  is 
§\%  of  your  money :  "  what  was  the  debt  ?  $160. 

16.  $25  is  62|%   of  A's   money,  and  41|^   of  B's:  how 
much  has  each?  A  $40,  B  $60. 

17.  A  found  $5,  which  was  13J%  of  what  he  had  before: 
how  much  had  he  then?  $42.50 

18.  I  drew  48%   of  my  funds  in  bank,  to  pay  a  note  of 
$150:  how  much  had  I  left?  $162.50 

19o  A  farmer  gave  his  daughter  at  her  marriage  65  A. 
106  sq.  rd.  of  land,  which  was  3^  of  his  farm  :  how  much 
land  did  he  own?  2188  A.  120  sq.  rd. 

20.  A  pays  $13  a  month  for  board,  which  is  20%  of  his 
salary  :  what  is  his  salary  ?  $780  a  year. 

21.  Paid  40  ct.  for  putting  in  25  bu.  of  coal,  which  was 
llf^  of  its  cost:  what  did  it  cost  a  bu.?  14  ct. 

22.  81  men  is  5%  of  60%  of  what?  2700  men. 
23o  A,  owning  60%  of  a  ship,  sells  7-|%  of  his  share  for 

$2500:  what  is  the  ship  worth?  $55555. 55f 

24.  A  father,  having  a  basket  of  apples,  took  out 
of  them  ;  of  these,  he  gave  37-^%  to  his  son,  who  gave 

of  his  share  to  his  sister,  who  thus  got  2  apples:  how  many 
apples  were  in  the  basket  at  first?  80  apples. 

25.  B  lost  three  dollars,   which  was  31^    of  what  he 
had  left:  how  much  had  he  at  first?  $12.60 

26.  Bought  8000  bu.  of  wheat,  which  was  57|^  of  my 
whole  stock:  how  much  had  I  before?  6000  bu. 

27.  If  32^  of  75%  of  800%  of  a  number  is  1539,  what 
is  that  number?  -  801  IT. 


PERCENTAGE.  195 


CASE    IV. 

239.  Given  the  rate  and  the  amount  or  the  differ- 
ence, to  find  the  base. 

PRINCIPLE.  —  The  base  is  equal  to  the  amount  divided  by  1 
plus  the  ratey  or  the  difference  divided  by  1  minus  the  rate. 

PROBLEM. — A  rents  a  house  for  $377,  which  is  an  advance 
of  16^  on  the  rent  of  last  year:  what  amount  did  he  pay 
last  year? 

OPERATION. 
$377-^-1.16  =  $325.00,  Ans. 

Or,    1  0  0  ft  =  rental  last  year ; 

16^)  =  increase  this  year  ; 
hence,  100^  +  16^  =  116J&=$377; 
1<&=$3.25; 

Ans. 


PROBLEM. — John    has    $136,    which    is    20^     less    than 
Joseph's  money:  how  many  dollars  has  Joseph? 

OPERATION. 
$136-i-(l  —  .2)=$170.00,  Ans. 

Or,    10  0  (jf0  —  Joseph's  money  ; 

—  John's  money  =  $136; 


.-.    1  0  0  (fc  —  $  1.7  X  1  0  0  =  $  1  7  0.0  0,  Ans. 
136;  i=$34;  and  f  =  $34  X  5  =  $1  7  0.00,  Ans. 


Rule. — Divide   the   sum   by  1    plus    the  rate,  or  divide  the 
difference  by  1  minus  the  rate;  the  quotient  will  be  the  base. 


196  RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 
• 

1.  $4.80  is  33$^g  more  than  what?  $3.60 

2.  |  is  50%  more  than  what?  f. 

3.  96  da.  is  lOO^g  more  than  what?  48  da. 

4.  2576  bu.  is  60%  less  than  what?  6440  bu. 

5.  87|  ct.  is  87!%  less  than  what?  $7. 

6.  42  mi.  60  rd.  is  55^  less  than  what?       93  mi.  240  rd. 

7.  2   Ib.    9|f   oz.    is    50%    less   than   what  number  of 
pounds?  5f§f  Ib. 

8.  T^  is  99|^  less  than  what?  155|. 

9.  $920.93f  is  337!%  more  than  what?  $210.50 

10.  $4358. 06^  is  233^%  more  than  what?  $1307. 41$ 

11.  In  64!    gal.    of  alcohol,   the    water    is    l^fo    of  the 
spirit:  how  many  gal.  of  each?  60  gal.  sp.,  4-J-  gal.  w. 

12.*  A  coat  cost  $32;  the  trimmings  cost  70%  less,  and 
the  making  50^  less,  than  the  cloth:  what  did  each  cost? 
Cloth  $17.77$,  trimmings  $5.33$,  making  $8.88f 

13.*  If  a  bushel  of  wheat  make  391-  Ib.  of  flour,  and  the 
cost  of  grinding  be  4%,  how  many  barrels  of  flour  can  a 
farmer  get  for  80  bu.  of  wheat?  15^  barrels. 

14.*  How  many  eagles,  each  containing  9  pwt.  16.2  gr.  of 
pure  gold,  can  I  get  for  455.6538  oz.  pure  gold  at  the  mint, 
allowing  \\%  for  expense  of  coinage?  928  eagles. 

15.  2047  is  10^  of  110%  less  than  what  number?     2300. 

16.  4246^  is   6%  of  50^    of   466|^    more  than   what 
number?  3725. 

17.  A  drew  out  of  bank  40  ^   of  50^  of  60%  of  70^ 
of  his  money,  and  had  left  $1557.20:  how  much  had  he  at 
first?  $1700. 

18.  I  gave  away  42-f  %  of  my  money,  and  had  left  $2 : 
what  had  I  at  first?  $3.50 

19.  In  a  school,  5%  of  the  pupils  are  always  absent,  and 
the  attendance  is  570 :    how   many  on   the   roll,    and  how 
many  absent?  600,  enrolled  ;  30,  absent. 


APPLICA  TIONS  OF  PER  CENT  A  GE.  197 

20.  A  man  dying,  left  33-|%  of  his  property  to  his  wife, 
60%  of  the  remainder  to  his  son,  75%  of  the  remainder  to 
his  daughter,  and  the  balance,  $500,  to  a  servant :   what  was 
the  whole  property,  and  each  share  ? 

Property,  $7500  ;  wife  had  $2500  ; 
son,  $3000 ;  daughter,  $1500. 

21.  In  a  company  of  87,  the  children  are  37^  of  the 
women,  who  are  44f%  of  the  men:  how  many  of  each? 

54  men,  24  women,  9  children. 

22.*  Our  stock  decreased  33|%,  and  again  20^  ;  then  it 
rose  20%,  and  again  33^^  ;  we  have  thus  lost  $66:  what 
was  the  stock  worth  at  first?  $450. 

23.*  A  brewery  is  worth  4%  less  than  a  tannery,  and 
the  tannery  16%  more  than  a  boat;  the  owner  of  the 
boat  has  traded  it  for  75  ^  of  the  brewery,  losing  thus 
$103:  what  is  the  tannery  worth?  $725. 


ADDITIONAL  FORMULAS. 

240.     The  following  additional  formulas,  derived   from 
preceding  data,  may  also  be  employed  to  advantage : 

1.  By  definition,  A  =  B  +  P ;  also,  D  =  B  —  P. 


2.  From  Case  IV.  < 


APPLICATIONS  OF  PERCENTAGE. 

241.  The  Applications  of  Percentage  may  be  divided 
into  two  classes,  those  in  which  time  is  not  an  essential 
element,  and  those  in  which  it  is  an  essential  element,  as 
follows : 


198 


RAY'S  HIGHER   ARITHMETIC. 


Those  in  which  Time  is  not  an 
Essential  Element 


1.  Profit  and  Loss. 

2.  Stocks  and  Dividends. 

3.  Premium  and  Discount. 

4.  Commission  and  Brokerage. 

5.  Stock  Investments. 

6.  Insurance. 

7.  Taxes. 

8.  United  States  Revenue. 

1.  Simple  Interest. 

2.  Partial  Payments. 

3.  True  Discount. 

4.  Bank  Discount. 

5.  Exchange. 

6.  Equation  of  Payments. 

7.  Settlement  of  Accounts. 

8.  Compound  Interest. 

9.  Annuities. 


NOTE. — These  topics  will  be  presented  in  the  order  in  which  they 
stand. 


Those  in  which  Time  is  an 

Essential  Element 


Topical  Outline. 
PERCENTAGE. 


J.  Definitions. 

2.  Notation. 

(  1.  Principle. 

I.       -j   2.  Formula. 

I  3.  Rules. 

3.  Cases  

f  1.  Principle. 
11.     -j   2.  Formula. 
I  3.  Rules. 

f  1.  Principle. 

III.   -j   2.  Formula. 

I  3.  Rules. 

C  1.  Principle. 

.  IV.    J   2.  Formula. 

I  3.  Rule. 

4.  Additional  Formulas. 

5.  Applications  of  Percentage.  /  Time 
I  Time 

not  an  Element, 
an  Element. 

XY.    PEKCENT AGE.  -APPLICATIONS. 

I.    PKOFIT  AND  LOSS. 
DEFINITIONS. 

242.  1.  Profit  and  Loss  are  commercial  terms,  and  pre- 
suppose a  cost  price. 

2.  The  Cost  is   the  price  paid  for  any  thing. 

3.  The  Selling  Price  is  the  price  received  for  whatever 
is  sold. 

4.  Profit  is  the  excess  of  the  Selling  Price  above  the  Cost. 

5.  Loss  is  the  excess  of  the  Cost  above  the  Selling  Price. 

243.  There  are  four  cases  of  Profit  or  Loss,  solved  like 
the  four  corresponding  cases  of  Percentage. 

The  cost  corresponds  to  the  Base;  the  per  cent  of  profit 
or  loss,  to  the  Rate;  the  profit  or  loss,  to  the  Percentage; 
the  cost  plus  the  profit,  or  the  selling  price,  to  the  Amount; 
and  the  cost  minus  the  loss,  or  the  selling  price,  to  the 
Difference. 

CASE    I. 

244.  Given  the  cost  and  the  rate,  to  find  the  profit 
or  loss. 

PROBLEM. — Having   invested    $4800, 

r»  r. ,     .       -I  o^y  -I      t     •  OPERATION. 

my  rate  of  profit  is  13%:   what  is  my  $4«00 

profit  ?  j  3 

SOLUTION. — Since   the   cost   is  $4800,  and  14400 

the  rate  of   profit  is  13$,;  the  profit  is  13$  4800 

of  $4800,  which  is  $624.  $  6  2  4.0  0   Profit. 

(199) 


200  RA  Y'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  If  a  man  invests  $1450  so  as  to  gain  14^%,  what  is 
his  profit?  $210.25 

2.  I   bought  $1760  worth  of  grain,  and  sold  it  so  as  to 
make  26£%  profit :  what  did  I  receive  for  it?  $2222. 

3.  If  a  man  invests  $42540,  and  loses  11|^  of  his  capital: 
to  what  does  his  loss  amount,   and  how  much   money  has 
he  left?  $4963,  loss;  $37577,  left. 

4.  A  man  buys   576  sheep,  at  $10  a  head.     If  his  flock 
increases  21^f  per  cent,  and  he  sells  it  at  the  same  rate  per 
head,  how  much  money  does  he  receive  ?  $7000. 

5.  The  cost  of  publishing  a  book  is  50  ct.  a  copy ;  if  the 
expense  of  sale  be  10%  of  this,  and  the  profit  25%  :  what 
does  it  sell  for  by  the  copy?  68f  ct. 

6.  A  began  business  with  $5000:  the  1st  year  he  gained 
14f  %,  which  he  added  to  his  capital ;  the  2d  year  he  gained 
8^,  which   he  added  to   his   capital;  the  3d   year  he  lost 
12%,  and  quit:  how  much  better  off  was  he  than  when  he 
started?  $452.92 

7.  A  bought   a   farm  of  government  land,  at   $1.25   an 
acre;  it  cost  him  160%  to  fence  it,  160%   to  break  it  up, 
80%  for  seed,  100%  to  plant  it,  100^  to  harvest  it,  112% 
for  threshing,  100^  for  transportation ;  each  acre  produced 
35  bu.    of  wheat,  which  he  sold  at  70   ct.   a  bushel :  how 
much  did  he  gain  on  every  acre  above  all  expenses  the  first 
year?  $13.10 

8.  For  what  must   I  sell  a  horse,  that  cost  me  $150,  to 
gain  35%?  $202.50 

9.  Bought  hams  at  8  ct.  a  lb.;  the  wastage  is  10% :  how 
must  I  sell  them  to  gain  30^  ?  llf  ct.  a  lb. 

10.  I  started  in  business  with  $10000,  and  gained  20% 
the  first  year,  and  added  it  to  what  I  had ;  the  2d  year  I 
gained   20%,  and  added  it  to   my  capital;  the   3d  year  I 
gained  20%  :  what  had  I  then?  $17280. 


PROFIT  AND  LOSS.  201 

11.  I  bought  a  cask  of  brandy,  containing  46  gal.,  at 
$2.50  per  gal.;  if  6  gal.  leak  out,  how  must  I  sell  the  rest, 
so  as  to  gain  25%  ?  $3.59f  per  gal. 

CASE  II. 

245.  Given  the  cost  and  the  profit  or  loss,  to  find 
the  rate. 

PROBLEM.  —  A  man  bought  part  of  a  mine  for  $45000, 
and  sold  it  for  $165000  :  how  many  per  cent  profit  did  he 
make? 

OPERATION. 
$165000  —  $45000  =  $120000  profit. 


1  ^  =  $  4  5  0;  and  120000-f-450--=266f^,  An*. 

SOLUTION.  —  Here    the  profit   is  $120000,  which,   compared   with 
$45000,  the  cost,  is  *$$££-  =  f  ,  that  is,  f  of  100^  =  266f  ft. 


EXAMPLES  FOR  PRACTICE. 

1.  If  I  buy  at  $1  and  sell  at  $4,  how  many  per  cent  do  I 
gain?  300%. 

2.  If  I  buy  at  $4  and  sell  at  $1,  how  many  per  cent  do 
Hose?  75%. 

3.  If  I  sell  |-  of  an  article  for  what  the  whole  cost  me, 
how  many  per  cent  do  I  gain  ?  80^ . 

4.*  Paid  8125  for  a  horse,  and  traded  him  for  another, 
giving  60^  additional  money.  For  the  second  horse  I  re- 
ceived a  third  and  $25  ;  I  then  sold  the  third  horse  for  $150: 
what  was  my  per  cent  of  profit  or  loss  ?  ^\%  loss. 

5.*  A  man  bought  a  farm  for  $1635,  which  depreciated  in 
value  25%.  Selling  out,  he  invested  the  proceeds  so  as  to 
make  33^  profit :  what  was  his  per  cent  of  profit  or  loss  on 
the  entire  transaction  ? 


202  RAY'S  HIGHER  ARITHMETIC. 

6.  It  cost  me  $1536  to  raise  my  wheat  crop:  if  I  sell  it 
for  $1728,  what  per  cent  profit  is  that  per  bushel?         12^%. 

7.  If  I  pay  for  a  Ib.  of  sugar,  and  get  a  Ib.  troy,  wrhat 
%    do  I  lose,  and  what   %    does   the  grocer  gain  by  the 
cheat?  17%%  loss;  21$$%  gain. 

8.  A,  having   failed,  pays    B  $1750  instead    of   $2500, 
which  he  owed  him:  what  %  does  B  lose?  30^. 

9.  An  article  has    lost    20^  by  wastage,  and  is  sold  for 
40%  above  cost:  what  is  the  gain  per  cent?  12%. 

10.  If  my  retail  profit    is    33^,  and    I   sell  at  whole- 
sale   for    10%    less   than  at  retail,    what  is  my  wholesale 
profit?  20%. 

11.  Bought  a  lot  of  glass;    lost  15%    by  breakage:  at 
what  %  above  cost  must  I  sell  the  remainder,  to  clear  20% 
on  the  whole?  ^TT%- 

12.  If  a  bushel  of  corn  is  worth  35  ct.  and  makes  2$  gal. 
of  whisky,  which   sells  at  $1.14  a  gal.,  what  is   the  profit 
of  the  distiller  who  pays  a  tax  of  90  ct.  a  gallon  ?         9T8^% . 

13.  I  had  a  horse  worth  $80 ;    sold  him   for  $90 ;  bought 
him  back  for  $100:  what  %  profit  or  loss?  ^\%  l°ss- 

CASE    III. 

246.  Given  the  profit  or  loss  and  the  rate,  to 
find  the  cost. 

PROBLEM. — By  selling  a  lot  for  34f  %  more  than  I  gave, 
my  profit  is  $423.50:  what  did  it  cost  me? 

SOLUTION. — Since  34f  <fc  =  $423.50,  OPERATION. 

\</0  =$423.50  -*-34f,  =  $12.32;    and  3  4  f  ft  =  $4  2  3.5  0; 
100^,  or  the  whole  cost,  =  100  times  1  fc  =  $  1  2.3  2 

$12.32  =  $1232,    as   in    Case    III    of  1  0  0  ft  =  $  1  2  3  2,  Ans. 
Percentage. 

REMARK. — After  the  cost  is  known,  the  profit  or  loss  may  be 
added  to  it,  or  subtracted  from  it,  to  get  the  selling  price  (amount 
or  difference). 


PROFIT  AND  L0 


EXAMPLES  FOR  PRACTICE. 

1.  How  large  sales  must  I  make  in  a  year,  at  a  profit  of 
8^,  to  clear  $2000?  $25000. 

2.  I  lost  $50  by  selling  sugar  at  22|^  below  cost  :  what 
was  the  cost?  $222.  22f 

3.  If  I  sell  tea  at  V&\%  profit,  I  make  10  ct.  a  lb.:  how 
much  a  pound  did  I  give?  75  ct. 

4.  I  lost  a  2|-  dollar  gold  coin,  which  was  1^%  of  all  I 
had:  how  much  had  I?  $35. 

5.  A  and   B   each  lost  $5,  which  was  2|^£   of  A's  and 
%\%  of  B's  money:  which  had  the  most  money,  and  how 
much  ?  A  had  $30  more  than  B. 

6.  I  gained  this  year  $2400,  which  is  120^  of  my  gain 
last  year,  and   that  is  44|-%   of  my  gain   the  year  before  : 
what  were  my  profits  the  two  previous  years  ? 

$2000  last  year;  $4500  year  before. 

7.  The  dogs  killed  40  of  my  sheep,  which  was  k\%  of  my 
flock  :  how  many  had  I  left  ?  920  sheep. 

CASE    IV. 

247.     Given  the  selling  price  (amount  or  difference) 
and  the  rate,  to  find  the  cost. 

PROBLEM.  —  Sold   goods   for   $25.80,  by  which    I   gained 
:  what  was  the  cost? 


SOLUTION.  —  The  cost  is  100^  ;  OPERATION. 

the   $25.80   being   7J^    more,  is  100^  =  cost  price  ; 

W7^/0  ;  then,  iyfl  =  $25.80  -=-  107  J  1  0  7  }  ft  =  $  2  5.8  0  ; 

=  24  ct.,  and  100^,,  or  the  cost,  —  /.  1^=24  ct. 

100  times   24  ct.  =  $24  ;    as    in  1  0  0  ft  =  $  2  4,  Ans. 

Case  IV  of  Percentage. 

EEMARK.  —  After  the  cost  is  found,  the  difference  between  it  and 
the  selling  price  (amount  or  difference)  will  be  the  profit  or  loss. 


204  RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  Sold  cloth  at  $3.85  a  yard;  my  profit  was  10^  :  how 
much  a  yard  did  I  pay?  $3.50 

2.  Gold   pens,  sold  at  $5  apiece,  yield  a  profit  of  33^%: 
what  did  they  cost  apiece?  $3.75 

3.  Sold  out  for  $952.82  and  lost  12%  :  what  was  the  cost, 
and  what  would  I  have  received  if  I  had  sold  out  at  a  profit 
of  12^  ?  $1082.75,  and  $1212.68 

4.  Sold    my  horse   at   40%    profit;   with  the  proceeds  I 
bought   another,  and  sold  him  for  $238,  losing  20%  :  what 
did  each  horse  cost  me?         $212.50  for  1st,  $297.50  for  2d. 

5.  Sold   flour   at   an    advance    of   13^^  ;     invested    the 
proceeds   in   flour    again,    and   sold   this   lot  at  a   profit  of 
24^,  realizing  $3952.50:  how  much  did  each  lot  cost  me? 

1st  lot,  $2812.50;  2d  lot,  $3187.50 

6.  An  invoice  of  goods  purchased  in  New  York,  cost  me 
8%  for  transportation,  and  I  sold  them  at  a  gain  of  16-|^ 
on  their  total   cost  on  delivery,   realizing   $1260 :  at  what 
were  they  invoiced?  $1000. 

7.  For  6  years  my  property  increased  each  year,  on   the 
previous,  100^ ,  and  became  worth  $100000 :  what   was  it 
worth  at  first?  $1562.50 


IT.    STOCKS  AND  BONDS. 
DEFINITIONS. 

248.  1.  A  Company  is  an  association  of  persons  united 
for  the  transaction  of  business. 

2.  A  company  is  called  a  Corporation  when  authorized 
by  law  to  transact  business  as  one  person. 

Corporations  are  regulated  by  general  laws  or  special 
acts,  called  Charters. 


STOCKS  AND  BONDS.  205 

3.  A  Charter  is  the  law  which  defines  the  powers,  rights, 
and  legal  obligations  of  a  corporation. 

4.  Stock    is    the    capital  of   the  corporation    invested  in 
business.     Those  owning  the  stock  are  Stockholders. 

5.  Stock   is  divided  into  Shares,  usually  of  $50   or  $100 
each. 

6.  Scrip  or  Certificates  of  Stock  are  the  papers  issued 
by  a  corporation  to  the  stockholders.     Each  stockholder  is 
entitled  to  certificates  showing  the  number  of  shares  that  he 
holds. 

7.  Stocks  is  a   general  term  applied  to  bonds,  state  and 
national,   and    to    the    certificates    of   stock    belonging    to 
corporations. 

8.  A  Bond  is  a  written  or  printed  obligation,  under  seal, 
securing  the    payment  of  a    certain   sum    of  money  at    or 
before  a  specified  time. 

Bonds  bear  a  fixed  rate  of  interest,  which  is  usually  payable 
either  annually  or  semi-annually. 

The  principal  classes  of  bonds  are  government,  state,  city, 
county,  and  railroad. 

9.  An  Assessment  is  a  sum  of  money   required   of  the 
stockholders  in  proportion  to  their  amounts  of  stock. 

REMARK. — Usually,  in  the  formation  of  a  company,  the  stock 
subscribed  is  not  all  paid  for  at  once ;  but  assessments  are  made  as  the 
needs  of  the  business  require.  The  stock  is  then  said  to  be  paid  for 
in  installments.  Other  assessments  may  be  made  to  meet  losses  or  to 

extend  the  business. 
i 

10.  A  Dividend   is  a  sum   of  money  to  be  paid  to  the 
stockholders  in  proportion  to  their  amounts  of  stock. 

REMARK. — The  gross  earnings  of  a  company  are  its  total  receipts  in 
the  transaction  of  the  business  ;  the  net  earnings  are  what  is  left  of  the 
receipts  after  deducting  all  expenses.  The  dividends  are  paid  out 
of  the  net  earnings. 


206  It  A  Y'S  HIGHER  ARITHMETIC. 

249.  Problems  involving  dividends  and  assessments  give 
rise  to  four  cases,  solved  like  the  four  corresponding  cases 
of  Percentage. 

The  quantities  involved  are,  the  Stock,  the  Rate,  and 
the  Dividend  or  Assessment. 

The  stock  corresponds  to  the  Base;  the  dividend  or 
assessment,  to  the  Percentage;  the  stock  plus  the  dividend, 
to  the  Amount;  and  the  stock  minus  the  assessment,  to 
the  Difference. 

CASE    I. 

250.  Given   the    stock    and   the    rate,    to    find    the 
dividend  or  assessment. 

FORMULA. — Stock  X  .Bate  =  Dividend  or  Assessment. 

EXAMPLES  FOR  PRACTICE. 

1.  I  own   18   shares,  of  $50  each,  in  the  City  Insurance 
Co.,  which   has  declared  a  dividend  of  1^%  •    what  do  I 
receive?  $67.50 

2.  I  own   147  shares   of  railroad    stock   ($50  each),  on 
which  I  am  entitled  to  a  dividend  of  5%,  payable  in  stock: 
how  many  additional  shares  do  I  receive  ? 

7  shares,  and  $17.50  toward  another  share. 

3.  The  Western  Stage  Co.  declares  a  dividend  of  4^  per 
cent:  if  their   whole  stock    is   $150000,  how   much  is   dis- 
tributed to  the  stockholders  ?  $6750. 

4.  The  Cincinnati  Gas  Co.  declares  a  dividend  of  18%  : 
what  do  I  get  on  50  shares  ($100  each)  ?  $900. 

5.  A  railroad  company,  whose  stock  account  is  $4256000, 
declared   a   dividend   of   3ffi  :  what   sum    was    distributed 
among  the  stockholders?  $148960. 

6.  A   telegraph   company,  with   a  capital  of  $75000,  de- 
clares a  dividend  of  7%,  and  has  $6500  surplus:   what  has 
it  earned?  $11750. 


STOCKS  AND  BONDS.  207 

7.  I  own  24  shares  of  stock  ($25  each)  in  a  fuel  company, 
which  declares  a  dividend  of  6%  ;  I  take  my  dividend  in 
coal,  at  8  ct.  a  bu. :  how  much  do  I  get  ?  450  bu. 

CASE  II. 

251.  Given  the  stock  and  dividend  or  assessment, 
to  find  the  rate. 

Dividend  or  Assessment       _. 

FORMULA. —  - —  =  Rate. 

Stock 


EXAMPLES  FOR  PRACTICE. 

1.  My  dividend  on  72  shares  of  bank  stock  ($50  each)  is 
$324:  what  was  the  rate  of  dividend?  9^. 

2.  A   turnpike   company,  whose  stock  is  $225000,  earns 
during  the  year  $16384.50:  what  rate  of  dividend  can  it 
declare?  7  ft,  and  $634.50  surplus. 

3.  The  receipts  of  a  certain  canal  company,  whose  stock 
is  $3650000,  amount    in   one    year   to    $256484;    the   out- 
lay is   $79383 :    what   rate   of  dividend   can   the    company 
declare?  ty%*  and  $12851  surplus. 

4.  I  own   500   shares  ($100)   in  a  stock  company.     If  I 
have  to  pay  $250  on  an  assessment,  what  is  the  rate?     \%* 

CASE    III. 

252.     Given   the    dividend    or    assessment    and    the 
rate,  to  find  the  stock. 

Dividend  or  Assessment        ~   * 

FORMULA. —  =  Stock* 

Rate 

EXAMPLES  FOR  PRACTICE. 

1.  An  insurance   company  earns  $18000,  and  declares  a 
15  ft  dividend:  what  is  its  stock  account?  $120000. 


208  RAY'S  HIGHER  ARITHMETIC. 

2  A   man  gets  $94.50   as   a   1°/0   dividend:    how  many 
shares  of  stock  ($50  each)  has  he?  27  shares. 

3  Keceived    5   shares    ($50   each),    and  $26   of  another 
share,  as    an    8%    dividend    on    stock:    how    many   shares 
had  I?  69  shares. 

CASE    IV. 

253.  Given  the  rate  and  the  stock  plus  the  divi- 
dend, or  the  stock  minus  the  assessment,  to  find  the 
stock. 

Stock  -f-  Dividend 

FORMULAS,-  Stock  =  1          l  +  Rate' 

Stock  —  Assessment 

I  —  Rate. 


EXAMPLES  FOR  PRACTICE. 

1.  Received  10^  stock  dividend,  and  then  had  102  shares 
($50  each),  and  $15  of  another  share:  how  many  shares  had 
I  before  the  dividend?  93  shares. 

2.  Having  received  two  dividends  in   stock,  one  of  5%, 
another  of  8  % ,  my  stock  has  increased  to  567  shares :  how 
many  had  I  at  first?  500  shares. 


III.    PREMIUM   AND  DISCOUNT. 

254.  1.  Premium,  Discount,  and  Par  are  mercantile 
terms  applied  to  money,  stocks,  bonds,  drafts,  etc. 

!2.  Drafts,  Bills  of  Exchange,  or  Checks  are  written 
orders  for  the  payment  of  money  at  some  definite  place  and 
time. 

3.  The  Par  Value  of  money,  stocks,  drafts,  etc.,  is  the 
nominal  value  on  their  face. 

4.  The  Market  Value  is  the  sum  for  which  they  sell. 


PREMIUM  AND  DISCOUNT.  209 

5.  Discount    is    the   excess  of  the   par  value  of  money, 
stocks,  drafts,  etc.,  over  their  market  value. 

6.  Premium  is   the  excess  of  their  market  value  over 
their  par  value. 

7.  Rate  of  Premium  or  Rate  of  Discount  is  the  rate 
per  cent  the  premium  or  discount  is  of  the  face. 

255.  Problems  involving  premium  or  discount  give  rise 
to  four  cases,  corresponding  to  those  of  Percentage. 

The  quantities  involved  are  the  Par  Value,  the  Rate, 
the  Premium  or  Discount,  and  the  Market  Value. 

The  par  value  corresponds  to  the  Base;  the  premium  or 
discount,  to  the  Percentage ;  and  the  market  value,  to  the 
Amount  or  Difference. 

CASE    I. 

256.  Given  the  par  value  and  the  rate,  to  find  the 
premium  or  discount. 


FORMULAS.-  {  P~m  =  Par  Value  X  Rate' 
(  Discount  =  Par  Value  X  -Rafe. 


NOTE. — If  the  result  is  a  premium,  it  must  be  added  to  the  par  to 
get  the  market  value ;  if  it  is  a  discount,  it  must  be  subtracted. 


EXAMPLES  FOR  PRACTICE. 

1.  Bought  54  shares  of  railroad  stock  ($100  each)  at  4% 
discount:   find  the  discount  and  cost. 

Dis.,  $216;  cost,  $5184 

2.  Buy  18  shares  of  stock  ($100  each)  at  8^   discount: 
find  the  discount  and  cost.  $144,  and  $1656. 

3.  Sell  the  same  at  4^-^  premium  :  find  the  premium,  the 
price,  and  the  gain.  $81,  $1881,  and  $225. 

4.  Bought  62  shares  of  railroad  stock  ($50  each)  at 


premium  :  what  did  they  cost?  $3968. 

H.  A.  18. 


210  RAY'S  HIGHER  ARITHMETIC. 

5.  What  is  the  cost  of  47  shares  of  railroad  stock  ($50 
each)  at  30%  discount?  $1645. 

6.  Bought    $150  in  gold,  at  \%   premium:  what  is  the 
premium  and  cost?  $1.12^,  and  $151.121 

7.  Sold  a  draft  on  New  "York  of  $2568.45,  at  \<fr  prem- 
ium :   what  do  I  get  for  it  ?  $2581.29 

8.  Sold  $425  uncurrent  money,  at  3%  discount:  what  did 
I  get,  and  lose?  $412.25,  and  $12.75 

9.  What  is  a  $5  note  worth,  at  6^  discount?          $4.70 

10.  Exchanged  32  shares  of  bank  stock  ($50  each),  5^ 
premium,  for  40  shares  of  railroad  stock  ($50  each),  10^  dis- 
count, and  paid  the  difference  in  cash  :  what  was  it?      $120. 

11.  Bought  98  shares  of  stock  ($50  each),  at  15%  dis- 
count ;  gave  in  payment  a  bill  of  exchange  on  New  Orleans 
for  $4000,  at  J- %   premium,  and  the  balance  in   cash:  how 
much  cash  did  I  pay?  $140. 

12.  Bought  56  shares   of  turnpike  stock  ($50  each),   at 
69%;  sold  them  at  1§\%  :  what  did  I  gain?  $210. 

13.  Bought  telegraph  stock   at   106%;   sold   it  at   91%: 
what  was  my  loss  on  84  shares  ($50  each)  ?  $630. 

14.  What  is  the  difference  between  a  draft  on  Philadel- 
phia of  $8651.40,  at  \\<fG  premium,  and  one  on  New  Orleans 
for  the  same  amount,  at  ^%  discount?  $151.40 

CASE     II. 

257.  Given  the  face  and  the  discount  or  premium, 
to  find  the  rate. 

Discount  or  Premium 

Par  Value. 

FORMULAS. — Rate  =  •<    r, .  „.          ,  .          ,  r    7  .       ,  T>     T7  / 
Difference  between  Market  and  Jrar   Value 

Par  Value. 

NOTES. — 1.  If  the  par  and  the  market  value  are  known,  take  their 
difference  for  the  discount  or  premium. 

2.  If  the  rate  of  profit  or  loss  is  required,  the  market  value  or  cost  is 
the  standard  of  comparison,  not  the  face. 


PREMIUM  AND  DISCOUNT.  211 


EXAMPLES  FOR  PRACTICE. 

1.  Paid   82401.30  for  a  draft  of  $2360  on  New  York: 
what  was  the  rate  of  premium?  ^-\%* 

2.  Bought  112  shares  of   railroad    stock  ($50  each)   for 
$3640  :  what  was  the  rate  of  discount?  35%. 

3.  If  the  stock  in  the  last  example  yields  8^  dividend, 
what  is  my  rate  of  profit?  12T4¥%. 

4.  I  sell  the  same  stock  for  $5936  :  what  rate  of  premium 
is  that?  what  rate  of  profit?  6%;  63T^. 

5.  If  I  count  my  dividend  as  part  of  the  profit,  what  is 
my  rate  of  profit  ?  75T5g^ . 

6.  Exchanged  12  Ohio  bonds  ($1000  each),  1%  premium, 
for  280  shares  of  railroad  stock   ($50  each)  :  what  rate  of 
discount  were  the  latter?  8^. 

7.  Gave  $266. 66|  of  notes,  4%  discount,  for  $250  of  gold: 
what  rate  of  premium  was  the  gold?  2-|%. 

8.  Bought  58  shares  of  mining  stock  ($50  each),  at  40% 
premium,  and  gave  in  payment  a  draft  on  Boston  for  $4000 : 
what  rate  of  premium  was  the  draft  ?  1  ^  % . 

9.  Received  $4.60  for  an  uncurrent  $5  note:  what  was 
the  rate  of  discount?  8%. 

10.  Paid  $2508.03  for  26  shares  of  stock  ($100  each),  and 
brokerage,  $25.03:  what  is  the  rate  of  discount? 

CASE    III. 

258.  Given  the  discount  or  premium  and  tlie  rate, 
to  find  the  face. 

.Discount  or  Premium 

FORMULA. —  Par  Value  — 

Rate. 

NOTES. — 1.  After  the  face  is  obtained,  add  to  it  the  premium,  or 
subtract  the  discount,  to  get  the  market  value  or  cost. 

2.  If  the  profit  or  loss  is  given,  and  the  rate  per  cent  of  the  face 
corresponding  to  it,  work  by  Case  III,  Percentage. 


212  RAY'S  HIGHER  ARITHMETIC, 


EXAMPLES  FOR  PRACTICE. 

1.  Paid  36   ct.   premium   for  gold  \%   above  par:  how 
much  gold  was  there?  $48. 

2.  Took  stock  at  par ;  sold  it  for  2ffi  discount,  and  lost 
$117:  how  many  shares  ($50  each)  had  I?  104  shares. 

3.  The   discount,  at   7^,   on  stocks,  was  $93.75:    how 
many  shares  ($50  each)  were  sold  ?  25  shares. 

4.  Buy  stock  at  4|%   premium;    sell  at  8^  premium; 
profit,  $345 :  how  many  shares  ($100  each)  ?  92  shares. 

5.  Buy  stocks  at  14^  discount;  sell  at   3^  premium; 
profit,  $192.50:  how  many  shares  ($50  each)?         22  shares. 

6.  The  premium   on  a  draft,   at  |^,  was  $10.36:  what 
was  the  face?  $1184. 

7.  Buy  stocks  at  6%  discount;  sell  at  42  ft  discount ;  loss, 
$666  :  how  many  shares  ($50  each)  ?  37  shares. 

8.  Bought    stock    at   10%    discount,    which   rose  to   5% 
premium,   and    sold    for    cash ;    paying  a    debt    of   $33,    I 
invested  the   balance   in   stock  at   2%  premium,  which,  at 
par,  left  me  $11   less  than  at  first:  how  much  money  had 
I  at  first?  $148.50 

CASE    IV. 

259.     Given  the  market  value  and  the  rate,  to  find 
the  par  value. 

Market  Value 


FORMULAS.- Par  Value  =  J    l  +  Eate  °f  Premium- 
]  Market  Value 

i    1  —  Eate  of  Discount. 

NOTES. — 1.  After  the  face  is  known,  take  the  difference  between  it 
and  the  market  value,  to  find  the  discount  or  premium. 

2.  Bear  in  mind  that  the  rate  of  premium  or  discount,  and  the 
rate  of  profit  or  loss,  are  entirely  different  things;  the  former  is 
referred  to  the  par  value  or  face,  as  a  standard  of  comparison,  the 
latter  to  the  market  value  or  cost. 


COMMISSION  AND  BROKERAGE.  213 

EXAMPLES  FOE,  PRACTICE. 

1.  What    is    the   face    of   a   draft   on  Baltimore   costing 
$2861.45,  at  \%<f0  premium?  $2819.16 

2.  Invested  $1591  in  stocks,  at  26^  discount :  how  many 
shares  ($50  each)  did  I  buy  ?  43  shares. 

3.  Bought  a   draft  on  New  Orleans,  at  -|%  discount,  for 
$6398.30:  what  was  its  face?  $6430.45 

4.  Notes  at  65%  discount,  1%   brokerage,  cost  $881.79: 
what  is  their  face?  $2470. 

5.  Exchanged  17  railroad  bonds  ($500  each)  25%  below 
par,   for   bank  stock  at   §\%  premium:    how  many  shares 
($100  each)  did  I  get?  60  shares. 

6.  How  much  gold,  at  f  %  premium,  will  pay  a  check  for 
$7567  ?  $7520. 

7.  How  much  silver,  at  1\%  discount,  can  be  bought  for 
$3172.64  of  currency  ?  $3212.80 

8.  How  large  a  draft,  at  \^0  premium,  is  worth  54  city 
bonds  ($100  each),  at  12%  discount?  $4740.15 

9.  Exchanged  72  Ohio  State  bonds  ($1000  each),  at  6^ 
premium,  for  Indiana  bonds  ($500  each),  at  2%  premium  : 
how  many  of  the  latter  did  I  get  ?  150  bonds. 


IV.    COMMISSION  AND  BROKEKAGE. 
DEFINITIONS. 

260.  1.  A  Commission-Merchant,  Agent,  or  Factor 
is  a  person  who  sells  property,  makes  investments,  collects 
debts,  or  transacts  other  business  for  another. 

2.  The  Principal  is  the  person  for  whom  the  commission- 
merchant  transacts  the  business. 

3.  Commission  is  the  percentage  paid  to  the  commission- 
merchant  for  doing  the  business. 


214  RA  Y>8  HIGHER  ARITHMETIC. 

4.  A  Consignment  is  a  quantity  of  merchandise  sent  to 
a  commission  merchant  to  be  sold. 

5.  The  person  sending  the  merchandise  is  the  Consignor 
or  Shipper,  and  the  commission  merchant  is  the  Consignee. 
When  living  at  a  distance  from  his  principal,  the  consignee 
is  spoken  of  as  the  Correspondent. 

6.  The  Net  Proceeds  is   the  sum  left  after  all  charges 
have  been  paid. 

7.  A  Guaranty  is  a  promise  to  answer  for  the  payment 
of  some  debt,  or  the  performance  of  some  duty  in  the  case 
of  the  failure  of  another  person,  who,  in  the  first  instance, 
is  liable.     Guaranties  are  of  two  kinds :  of  payment,  and  of 
collection. 

8.  In  a  Guaranty  of  Payment,  the  guarantor  makes  an 
absolute  agreement  that  the  instrument  shall   be  paid    at 
maturity. 

9.  The  usual  form  of  a  guaranty,  written  on  the  back  of  a 
note  or  bill,  is: 

"-For  value  received,  I  hereby  guaranty  the  payment  of  the 
within.  JOHN  SAUNDERS." 

10.  A  Broker  is  a  person  who  deals  in  money,  bills  of 
credit,  stocks,  or  real  estate,  etc. 

11.  The  commission  paid  to  a  broker  is  called  Brokerage. 

281.  Commission  and  Brokerage  involve  four  cases, 
corresponding  to  those  of  Percentage. 

The  quantities  involved  are  the  Amount  Bought  or  Sold, 
the  Rate  of  Commission  or  Brokerage,  the  Commission  or  Bro- 
kerage, and  the  Cost  or  Net  Proceeds. 

The  amount  of  sale  or  purchase  corresponds  to  the  Base; 
the  commission  or  brokerage,  to  the  Percentage;  the  cost,  to 
the  Amount;  and  the  net  proceeds,  to  the  Difference. 


COMMISSION  AND  BROKERAGE.  215 


CASE    I. 

262.     Given  the   amount   of  sale,  purchase,  or  col- 
lection and  the  rate,  to  find  the  commission. 


FORMULA.  —  Amount  of  Sale  or  Purchase  X  Rate  =  Commission. 


EXAMPLES  FOR  PRACTICE. 

1.  I  collect  for  A  $268.40,  and  have   5%    commission: 
what  does  A  get?  $254.98 

2.  I  sell  for  B  650  barrels  of  flour,  at  $7.50  a  barrel,  28 
barrels  of  whisky,   35   gal.  each,  at  $1.25  a  gal.:    what  is 
my  commission,  at  2^%?  $137.25 

3.  Received  on  commission  25  hhd.  sugar  (36547  lb.),  of 
which  I  sold  10  hhd.  (16875  lb.),  at  6  ct.  a  lb.,  and  6  hhd. 
(8246  lb.)  at  5  ct.  a  lb.,  and  the  rest  at  5^-  ct.  a  lb.:  what 
is  my  commission,  at  3%?  $61.60 

4.  A  lawyer  charged  8^  for  collecting  a  note  of  $648.75: 
what  are  his  fee  and  the  net  proceeds? 

$51.90,  and  $596.85 

5.  A  lawyer,  having  a  debt  of  $1346.50  to  collect,  com- 
promises by  taking  80%,  and  charges  5%  for  his  fee:  what 
are  his  fee,  and  the  net  proceeds?          $53.86,  and  $1023.34 

6.  Bought  for  C,  a  carriage  for  $950,  a  pair  of  horses  for 
$575,  and  harness  for  $120;  paid  charges  for  keeping,  pack- 
ing, shipping,  etc.,  $18.25;    freight,  $36.50:  what  was  my 
commission,  at  3-|^ ,  and  what  was  the  whole  amount  of  my 
bill?  $54.83,  and  $1754.58 

7.  An  architect   charges  3^^   for  designing  and  superin- 
tending a  building,  which  cost  $27814.60:  to  what  does  his 
fee  amount?  $973.51 

8.  A   factor    has    2|^    commission,  and    %\%   for   guar- 
antying  payment:    if  the   sales  are   $6231.25,    what  does 
he  get?  $389.45 


216  RAY'S  HIGHER  ARITHMETIC. 

9.  Sold  500000  Ib.  of  pork  at  5£  ct.   a  lb.:  what  is  my 
commission  at  \\%  ?  $343.75 

10.  An  architect  charges  \\%  f°r  plans  and  specifications, 
and   2J^    for    superintending:    what  does  he  make,  if  the 
building  costs  $14902.50  ?  $614.73 

11.  A  sells  a  house  and  lot  for  me  at  $3850,  and  charges 
%%  brokerage:  what  is  his  fee?  $24.06+ 

12.  I  have   a  lot  of  tobacco  on  commission,  and   sell   it 
through  a  broker  for  $4642.85:   my  commission  is  2^%,  the 
brokerage  \\%\  what  do  I  pay  the  broker,  and  what  do  I 
keep?  I  keep  $63.84;  brokerage,  $52.23 

CASE    II. 

263.     Given  the  commission  and  the  amount  of  the 
sale,  purchase,  or  collection,  to  find  the  rate. 

Commission 

FORMULA.— — —   =  Rate. 

Amount  of  oale  or  Purchase 


EXAMPLES  FOR  PRACTICE. 

1.  An  auctioneer's   commission  for  selling  a  lot  was  $50, 
and  the  sum  paid  the  owner  was  $1200 :  what  was  the  rate 
of  commission  ?  4  %  - 

2.  A  commission-merchant   sells  800  barrels  of  flour,   at 
$6.43|  a  barrel,    and  remits    the    net   proceeds,   $5021.25: 
what  is  his  rate  of  commission?  ^\%> 

3.  The  cost   of  a  building  was  J19017.92,  including  the 
architect's   commission,   which   was   $553.92:  what  rate  did 
the  architect  charge?  3  ft. 

4.  Bought  flour  for  A;  my  whole  bill  was  $5802.57,  in- 
cluding charges,  $76.85,  and  commission,  $148.72:  find  the 
rate  of  commission.  .      2f  %. 

5.  Charged  $52.50  for  collecting  a  debt  of  $1050:  what 
was  my  rate  of  commission  ?  5%, 


COMMISSION  AND  BROKERAGE.  217 

6.  An  agent  gets  $169.20  for  selling  property  for  $8460: 
what  was  his  rate  of  brokerage?  2%. 

7.  My  commission  for  selling  books  was  $6.92,  and  the  net 
proceeds,  $62.28:   what  rate  did  I  charge?  10^. 

8.  Paid  $38.40  for  selling  goods  worth  $6400:  what  was 
the  rate  of  brokerage  ?  f  % . 

9.  Paid  a  broker  $24.16,  and  retained  as  my  part  of  the 
commission  $42.28,  for  selling  a  consignment  at  $2416:  what 
was  the  rate  of  brokerage,  and  my  rate  of  commission  ? 

Brok.  1%;  com.  2f^. 

CASE    III. 

264.  Given  the  commission  and  rate,  to  find  the 
sum  on  which  commission  is  charged. 

Commission        .  ~  ,         n      . 

FORMULA. =  Amount  of  Sale  or  Purchase. 

Rate 

NOTE. — After  finding  the  sura  on  which  commission  is  charged, 
subtract  the  commission  to  find  the  net  proceeds,  or  add  it  to  find 
the  whole  cost,  as  the  case  may  be. 


EXAMPLES  FOR  PRACTICE. 


1.  My  commissions   in  1  year,  at  2^%,  are  $3500:  what 
were  the  sales,  and  the  whole  net  proceeds? 

8140000,  and  $136500. 

2.  An  insurance  agent's  income  is  $1733.45,  being  10^ 
on  the  sums  received  for  the  company:  what  were  the  com- 
pany's net  receipts?  $15601.05 

3.  A  packing-house  charged  \\%  commission,  and  cleared 
32376.15,   after  paying  out  $1206.75   for  all   expenses  of 
packing:  how  many  pounds  of  pork  were  packed,  if  it  cost 
41  ct.  a  pound?  5308000  Ib. 

4.  Paid  $64.05  for  selling  coffee,  which  was  \<fa  broker- 
age: what  are  the  net  proceeds?  $7255.95 

H.  A.  19. 


218  RAY'S  HIGHER  ARITHMETIC. 

5.  An  agent  purchased,  according  to  order,  10400  bushels 
of  wheat ;  his  commission,  at  1  \^0 ,  was  $156,  and  charges 
for  storage,  shipping,  and  freight,  $527.10:  what  did  he  pay 
a  bushel,  and  what  was  the  whole  cost  ? 

$1.20  a  bu.,  and  $13163.10,  whole  cost, 

6.  Received  produce  on  commission,  at  ^\%\  my  surplus 
commission,  after  paying  \<fo    brokerage,  is  $107.03:  what 
was   the  amount  of  the  sale,  the  brokerage,  and  net  pro- 
ceeds? Sale,  $6116;  brok.,  $30.58;  pro.,  $5978.39 

CASE   IV. 

265.  Given  the  rate  of  commission  and  the  net 
proceeds  or  the  whole  cost,  to  find  the  sum  on  which 
commission  is  charged. 

•  =  Amount  of  Sale  or  Purchase. 


_,  ,    (I  +  Rate) 

FORMULAS.—  -<    \r  '  „       ' 

Net  Proceeds  _      . 

—  =  Amount  of  Sale  or  Purchase. 
(I  —  Rale) 

NOTE. — After  the  sum  on  which  commission  is  charged  is  known, 
find  the  commission  by  subtraction. 


EXAMPLES  FOR  PRACTICE. 

1.  A  lawyer  collects  a  debt  for  a  client,  takes  4^  for  his 
fee,   and  remits  the  balance,   $207.60:  what  was  the   debt 
and  the  fee?  $216.25,  and  $8.65 

2.  Sent  $1000  to  buy  a  carriage,  commission  2^^  :  what 
must  the  carriage  cost?  $975.61 

SUGGESTION.— 100^  +  2J-^  =  102J$>  =  $1000  ;  find  1^,  then  100^0. 

3.  A  buys  per  order   a  lot  of  coffee;    charges,   $56.85; 
commission,  \\%\  the' whole  cost  is  $539.61:  what  did  the 
coffee  cost?  $476.80 


COMMISSION  AND  BROKERAGE.  219 


4.  Buy  sugar  at  2|^   commission,  arid  %^%  for  guaran- 
teeing payment:    if  the  whole  cost  is  $1500,  what  was  the 
cost  of  the  sugar?  $1431.98 

5.  Sold  2000  hams  (20672  Ib.)  ;  commission,  2^,  guar- 
anty, 2|^,  net  proceeds  due  consignor,  $2448.34:  what  did 
the  hams  sell  for  a  Ib.?  12|  ct. 

6.  Sold  cotton  on  commission,  at   5^;  invested    the  net 
proceeds  in  sugar;  commission,  2%;  my  whole  commission 
was  $210:  what  was  the  value  of  the  cotton  and  sugar? 

Cotton,  $3060;  sugar,  $2850. 

SUGGESTION.  —  Note  carefully  the  different  processes  required  here 
for  commission  in  buying  and  commission  in  selling. 


7.  Sold    flour    at    3|%    commission;     invested   f   of    its 
value  in  coffee,  at  l^/^  commission;    remitted  the  balance, 
$432.50:  what   was  the  value   of  the  flour,  the  coffee,  and 
my  commissions?  Flour,  $1500;  coffee,  $1000,  1st 

com.,  $52.50,  2d  com.,  $15. 

8.  Sold  a  consignment  of  pork,  and  invested  the  proceeds 
in  brandy,  after  deducting  my  commissions,  4%  for  selling, 
and  \\<fa    for   buying.     The    brandy  cost    $2304.00  :    what 
did  the  pork  sell  for,  and  what  were  my  commissions? 

Pork,  $2430;  1st  com.,  $97.20;  2d  com.,  $28.80 

9.  Sold  1400  barrels  of  flour,  at  $6.20  a  barrel;  invested 
the  proceeds  in  sugar,  as  per  order,  reserving  my  commis- 
sions,  4%    for  selling    and    Ig^J?    for  buying,  and   the  ex- 
pense   of  shipping,    $34.16:    how    much    did    I    invest    in 
sugar?  $8176. 

10.  An  agent  sold  my  corn;  and,  after  reserving  his  com- 
mission, invested  all  the  proceeds  in  corn  at  the  same  price  ; 
his  commission,  buying  and  selling,  was  3%,  and  his  whole 
charge  $12  :  for  what  was  the  corn  first  sold  ?  $206. 

11.  My  agent  sold  my  flour  at  4%  commission  ;  increas- 
ing the  proceeds  by  $4.20,  I  ordered  the  purchase  of  wheat 
at   2%   commission;  after  which,  wheat  declining  3J%,  my 
whole  loss  was  $5:  what  was  the  flour  worth?  $53. 


220  EAY'S  HIGHER  ARITHMETIC. 

V.    STOCK   INVESTMENTS. 
DEFINITIONS. 

266.  1.  A  Stock  Exchange  is  an  association  of  brokers 
and  dealers  in  stocks,  bonds,  and  other  securities. 

REMARKS. — 1.  The  name  "Stock  Exchange"  is  also  applied  to 
the  building  in  which  the  association  meets  to  transact  business. 

2.  New  York  city  is  the  commercial  center  of  the  United  States, 
and  the  transactions  of  the  New  York  Stock  Exchange,  as  tele- 
graphed throughout  the  country,  determine  the  market  value  of 
nearly  all  stocks  sold. 

2.  United  States  Government  Bonds  are  of  two  kinds, — 
coupon  and  registered. 

REMARKS.—  -1.  Coupon  bonds  may  be  transferred  like  bank-notes: 
the  interest  is  represented  by  certificates,  called  coupons,  printed  at 
the  bottom  of  the  bond,  which  may  be  presented  for  payment  when 
due. 

2.  Registered  bonds  are  recorded  in  the  name  of  the  owner  in  the 
U.  S.  Treasurer's  office,  and  the  interest  is  sent  directly  to  the  owner. 
Registered  bonds  must  be  indorsed,  and  the  record  must  be  changed, 
to  effect  a  transfer. 

3.  The   United   States    also    issues   legal  tender  notes,  known    as 
"  Greenbacks,"  which  are  payable  in  coin  on  demand,  and  bear  no 
interest. 

3.  The   various    kinds  of  United   States   bonds   are    dis- 
tinguished, 1st.  By  the  rate  of  interest;  2d.  By  the  date 
at  which  they  are  payable.     Most  of  the  bonds  are  payable 
in  coin ;  a  few  are  payable  in  currency. 

EXAMPLE. — Thus,  "U.  S.  4J's,  1891,"  means  bonds  bearing  in- 
terest at  4J^,,  and  payable  in  1891.  "  U.  S.  cur.  6's,  1899,"  means 
bonds  bearing  interest  at  6^,  and  payable  in  currency  in  1899. 
Quotations  of  the  principal  bonds  are  given  in  the  leading  daily 
papers. 

4.  Bonds  are  also   issued  by  the  several  states,  by  cities 
and  towns,  by  counties,  and  by  corporations. 


STOCK  INVESTMENTS.  221 

REMARKS. — Legitimate  stock  transactions  involve  the  following 
terms  and  abbreviations,  which  need  explanation  : 

1.  A  person  who   anticipates  a  decline,  and  contracts  to  deliver 
stocks  at  a  future  day,  at   a  fixed   price  which  is  lower  than  the 
present  market  price,  expecting  to  buy  in  the  interval  at  a  still 
lower  price,  is  said  to  sell  short. 

Short  sales  are  also  made  for  cash,  deliverable  on  the  same  day, 
or  in  the  regular  way,  where  the  certificates  are  delivered  the  day 
after  the  sale.  In  these  cases  the  seller  borrows  the  stock  from  a 
third  party,  advancing  security  equivalent  to  the  market  price,  and 
waits  a  decline;  he  buys  at  what  he  considers  the  lowest  point, 
returns  his  borrowed  stock,  and  reclaims  his  security. 

2.  A   person  who  buys  stock  in  anticipation  of  a  rise  is  said  to 
buy  long. 

3.  Those  who  sell  short   are    interested,   of    course,    in   forcing 
the   market   price   down,  and   are    called    bears;    while   those    who 
buy  long    endeavor  to  force  the  market  price  up,  and  are  known 
as  bulls. 

4.  The  following  are   the  principal   abbreviations  met   with  in 
stock  quotations:  c.,  means  coupon;  r.,  registered;  prefd,  or  pf.,  pre- 
ferred, applied  to  stock  which  has  advantages  over  common  stock 
of   the  same  company  in   the  way  of  dividends,  etc.;  xd.,  without 
dividend,  meaning  that  the  buyer  is  not  entitled  to   the  dividend 
about  to  be  declared;  c.,  cash;  s3,  s30,  s60,  seller's  option,  three,  thirty, 
or  sixty  days,  as  the  case  may  be,  means  that  the  seller  has  the  privi- 
lege  of  closing  the   transaction  at  any  time  within  the  specified 
limit;    b3,  b30,   b60,  buyer's  option,  three  days,  etc.,  giving  the  buyer 
the   privilege;    be.,  between  calls,   means    that   the   price   was   fixed 
between  the  calls  of  the  whole  list  of   stocks,  which  takes  place  in 
the  New  York  Exchange  twice  a  day  ;  opg.,  for  delivery  at  the  opening 
of  the  books  of  transfer. 

5.  The  usual  rate  of  brokerage   is   \<jc  on  the  par  value  of  the 
stock,  either  for  a  purchase  or  a  sale. 

267.  The  quantities  involved  in  problems  in  stock  in- 
vestments are :  the  Amount  Invested,  the  Market  Value,  the 
Kate  of  Dividend  or  Interest  paid  on  the  par  value  of  the 
stock,  the  Rate  of  Income  on  the  amount  invested,  and  the 
Income  itself. 

These  quantities  give  rise  to  five  cases,  all  of  which  may 
be  solved  by  the  principles  of  Percentage. 


,222  RAY'S  HIGHER  ARITHMETIC. 


NOTATION. 

268.  The    following   notation    may*  be    adopted   to   ad- 
vantage in  the  formulas : 

Amount  Invested  =  A.    I. 

Market  Value  =  M.  V. 

Rate  of  Dividend  or  Interest  =  R.  D. 

Rate  of  Income  =  R.    I. 

Income  =  I. 

CASE    I. 

269.  Given  the  amount  invested,  the  market  value, 
and  the  rate  of  dividend   or  interest,  to  find   the   in- 
come. 

A    I 

FORMULA.—  ~^— :  X  H.  B.  =  T 

PROBLEM. — A  person  invests  $5652.50  in  Mutual  In- 
surance Company  stock  at  95  cents :  what  will  be  his 
income  if  .the  stock  pay  10%  dividend  annually? 

OPERATION. 

$ 5  6  5  2 . 5  0>f  ,9  5  ==: $ 5 9  5  0  =  par  value  of  stock  purchased. 

$5950X-1=$595  —  income. 
Or,  9  5  ft  =  if  of  the  par  value  =  $5652.50 
5^  =  ^   "     rt       "         "     =$    297.50 
1  0  $  =  &   "     "       "         "      =  $    5  9  5,  Arts. 

SOLUTION. — As  many  dollars'  worth  of  stock  can  be  bought  as 
$.95  is  contained  times  in  $5652.50,  which  is  5950  times.  Therefore 
$5950  is  the  par  value  of  the  stock  purchased ;  and  10^  dividend  on 
$5950  is  $595,  the  income  on  the  investment. 


EXAMPLES  FOR  PRACTICE. 

1.  A  invests   $28000   in  Lake   Shore  Railroad   stock,  at 
If  the  stock  yields  8^  annually,  what  is  the  amount 
of  his  income?  $3200. 


STOCK  INVESTMENTS.  223 

2.  B  invests  $100962  in  U.  S.  cur.  6's,  1899,  at  106^%. 
If  gold  is  at  |-  premium,  what   does   the   government   save 
by  paying  him  his  interest  in  greenbacks?  $7.11 

3.  If   I  invest  $10200  in  Tennessee  6's,   new,  at  30%, 
what  is  my  annual  income  ?  $2040. 

4.  A    broker   invested    $36000    in    quicksilver    preferred 
stock,  at  40  ^  :    if  the  stock  pays  4^,  what  is  the  income 
derived?  $3600. 

5.  Which    is   the  better  investment,  stock    paying 
dividend,  at  a  market  value  of  106^%,  or  stock  paying 
dividend,  at  104%%  ?  The  former,  l^fJJ^. 

6.  Which  is  the  more  profitable,  to  invest  $10000  in  6% 
stock  purchased  at  75%,  or  in  5^  stock  purchased  at  60%, 
allowing  brokerage  \%  ?  5^  stock  is  $31.74+  better. 

CASE    II. 

270.  Given  the  amount  invested,  the  market  value, 
and  the  income,  to  find  the  rate  of  dividend  or  in- 
terest. 

A.  I. 

FORMULA.  —  I.  -*-  -     -7-  —  K.  D. 
M.  V. 

PROBLEM.  —  Invested  $10132.50  in  railroad  stock  at  105%, 
which  pays  me  annually  $965  :  what  is  the  rale  of  dividend 
on  the  stock? 

SOLUTION.—  By  Art.  259,  $10132.50  -*-  1.05  =  $9650  =  par  value 
of  the  stock  ;  and  by  Art.  251,  $965  -r-  $9650  =  .1,  or  10^0,  Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  A  has  a  farm,  valued  at  $46000,  which  pays  him  5% 
on  the  investment.  Through  a  broker,  who  charges  $56.50 
for  his  services,  he  exchanges  it  for  insurance  stock  at  9^ 
premium,  and  this  increases  his  annual  income  by  $1072: 
what  dividend  does  the  stock  pay?  8%. 


224  RAY'S  HIGHER   ARITHMETIC. 

2.  What  dividend  must  stock  pay,  in  order  that  my  rate 
of  income   on  an   investment  of  $64968.75  shall  be 
provided  the  stock  can  be  bought  at  103^%  ? 

3.  My   investment   was  $9850,   my  income  is  $500,   and 
the  market  value  of  the  stock  108^%,  brokerage  \%  :  what 
is  the  rate  of  dividend  ?  §\%  • 

4.  The  sale   of  my  farm   cost  me  $500,  but  I  gave  the 
proceeds  to  a  broker,  allowing  him  \%,  to  purchase  railroad 
stock  then  in  market  at  102%  ;  the  farm  paid  a  5^  income, 
equal  to  $2075,  but  the  stock  will  pay  $2025  more :  what  is 
the  rate  of  dividend?  10^%. 

5.  Howard  has  at   order  $122400,  and  can  allow  broker- 
age ^%,and  buy  insurance  stock  at  101^,  yielding  k\%\ 
but  if  he  send  to  the  broker  $100  more  for  investment,  and 
buy  rolling-mill  stock  at  103^ ,  the  income  will  only  be  half 
so  large:  what  rate  does  the  higher  stock  pay?  2¥6^. 

CASE   III. 

271.  Given  the  income,  rate  of  dividend,  and  mar- 
ket value,  to  find  the  amount  invested. 

FORMULA.—  -^-  X  M.  V.  -  A.  I. 

Jti.    1). 

PROBLEM. — If  U.  S.  bonds,,  paying  5%  interest,  are  sell- 
ing at  108^%,  how  much  must  be  invested  to  secure  an 
annual  income  of  $2000  ? 

OPERATION. 

$2000  —  $. 05  =  40000;  $40  000  X  1  -08  5  =$43  4  00,  Ans. 

SOLUTIONS. — 1.  To  produce  an  income  of  $2000  it  will  require  as 
many  dollar's  worth  of  stock  at  par  as  5  ct.  is  contained  times  in 
$2000,  which  is  40000;  and,  atlOSJ^,  it  will  require  $40000  X  1.08 \ 
=  $43400. 

2.  $1  of  stock  will  give  5  cents  income,  and  $2000  income  will 
require  $40000  worth  of  stock  at  par.     $40000  of  stock,  at 
will  cost  $40000  X  1-085  =  $43400. 


STOCK  INVESTMENTS.  225 


EXAMPLES  FOR  PRACTICE. 

1.  What  amount  is  invested   by  A,  whose   canal  stock, 
yielding  4^,  brings  an  income  of  $300,  but  sells  in  market 
for  92^  ?  $6900. 

2.  If  I  invest  all  my  money  in  5^  furnace  stock,  salable 
at  75%,  my  income  will  be  $180:  how  much  must  I  borrow 
to  make  an  investment  in  6^  state  stock,  selling  at  102^ , 
to  have  that  income?  $360. 

3.  If  railroad  stock  be  yielding  6^,  and  is  20%  below  par, 
how  much  would  have  to  be  invested  to  bring  an  income 
of  $390?  $5200. 

4.  A  banker  owns  2-|%  stocks,  at   10%  below   par,  and 
3%  stocks,  at  15%  below  par.     The  income  from  the  former 
is  66|^   more  than  from  the  Tatter,  and  the  investment  in 
the  latter   is  $11400  less  than  in  the  former:  required  the 
whole  investment  and  income.  $31800,  and  $960. 

5.  Howard  M.  Holden  sold  $21600  U.  S.  4's,  1907,  reg- 
istered, at   99|^,    and   immediately    invested    a   sufficient 
amount  of  the  proceeds  in  Illinois  Central  Railroad  stock, 
at  80%,  which  pays  an  annual  dividend  of  6^;  he  receives 
$840  from   the  railroad  investment ;  with  the  remainder  of 
his  money  he  bought  a  farm  at  $30  an  acre:  required  the 
amount  invested  in  railroad  stock,  and  the  number  of  acres 
in  the  farm?  $11200  in  R.  R.  stock;  342£  acres. 

6.  W.  T.  Baird,  through  his  broker,   invested   a   certain 
sum   of  money   in   Philadelphia   6's,   at   115^,  and   three 
times  as   much  in  Union  Pacific    7's,   at   89^%,   brokerage 
\%  in  both  cases :  how  much  was  invested  in  each  kind  of 
stock  if  his  annual  income  is  $9920? 

$34800  in  Phila.  6's ;  $104400  in  U.  P.  7's. 

7.  Thomas  Reed,  bought  6%  mining  stock  at  114^%,  and 
4^£  furnace  stock  at  112%,  brokerage  \%\  the  latter  cost  him 
$430  more  than  the  former,  but  yielded  the  same  income: 
what  did  each  cost  him?  Mining,  $920;  furnace,  $1350. 


226  RA  Y'  S  HIGHER  ARITHMETIC. 


CASE    IV. 

272.  Given  the  market  value  and  the  rate  of  divi- 
dend or  interest,  to  find  the  rate  of  income. 

T?    T) 

FORMULA. —  rr  -~=E.  I. 
M.  V. 

PROBLEM. — What  per  cent  of  his  money  will  a  man 
realize  in  buying  6%  stock  at  80%? 

OPERATION. 
$.0*4-$.SO  =  A  =  &  =  71£ 

SOLUTION. — The  expenditure  of  80  ct.  buys  a  dollar's  worth  of 
stock,  giving  an  income  of  6  ct.  The  rate  per  cent  of  income  on 
investment  is  $.06  -r-  $.80  —  .07J,  or  7J< 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  rate  of  income  on  Pacific  Mail  6's,  bought 
at  30^?  20%. 

2.  What  is  the  rate  of  income  on  Union  Pacific  6's,  bought 
at  110%?  5^. 

3.  Which  is  the  better  investment:  U.  S.  new  4's,  regis- 
tered, at  99|^;  or  U.  S.  new  4^'s,  coupons,  at  106%  ? 

The  latter  is  -f-f-^yo  better. 

4.  Thomas    Sparkler    has    an    opportunity    of   investing 
$30000  in  North-western   preferred    stock,  at   76%,   which 
pays  an  annual  premium  of  5^;  in  Panama  stock,  at  125^, 
which   pays  a  premium  annually  of  8^%;  or  he   can  lend 
his  money,  on  safe  security,   at   6|%  per  annum.     Prove 
which  is  the  best  investment  for  Mr.  Sparkler. 

The  Panama  stock. 

5.  Thomas  Jackson  bought  500  shares  of  Adams  Express 
stock,  at  105|^,  and  paid  1^  brokerage :  what  is  the  rate 
of  income  on  his  investment  per  annum  if  the  annual  divi- 
dend is  8%? 


STOCK  INVESTMENTS.  227 


CASE    V. 

273.     Given   the    rate    of  income    and   the   rate   of 
dividend  or  interest,  to  find  the  market  value. 

FOHMULA. — — -  — '  —  M.  V. 
K.  I. 

PROBLEM. — What  must  I  pay  for  Lake  Shore  6's  ($100  a 
share),  that  the  investment  may  yield  10^  ? 

OPERATION. 

a  share. 


SOLUTION. — If  bought  for  $100,  or  par,  it  will  yield  6^> ;  to  yield 
l<fo  it  must  be  bought  for  $100  X  6  =  $600  ;  to  yield  10^  it  must  be 
bought  for  yV  of  $600  =  $60. 


EXAMPLES  FOR  PRACTICE. 

1.  What  must  I  pay  for  Chicago,  Burlington  &  Quincy 
Railroad  stock   that  bears   6^,  that  my  annual  income  on 
the  investment  may  yield  5%  ?  120^. 

2.  Which  is  the  best  permanent  investment :  4's  at  70^ , 
5's  at  80%,  6's  at  90%,  or  10's  at  120^  ?     Why? 

3.  *The  rate  of  income  being  7%  on  the  investment,  and 
the  dividend  rate  4%,  what  is  the  market  value  of  $3430  of 
the  stock?  $1960. 

4.  In  a  mutual  insurance  company  one  capitalist  has  an 
investment  paying  8%:  what  is  the  premium  on  the  stock, 
the  dividend  being  9^  ?  ^\%> 

5.  Suppose  10%  state  stock  20%  better  in  market  than 
4%   railroad  stock;  if  A's   income  be  $500  from  each,  how 
much  money  has  he  'paid   for  each,   the  whole  investment 
bringing  &^%  ?  $11250,  railroad  ;  $5400,  state. 

6.  At  what  figure   must  be   government  5  per  cent's  to 
make  my  purchase  pay  9%? 


228  RA  Y>  #  HIGHER  ARITHMETIC. 

VI.    INSURANCE. 
DEFINITIONS. 

274.     1.  Insurance  is  indemnity  against  loss  or  damage. 

2.  There  are  two  kinds  of   insurance,  viz.:    Property  In* 
surance  and  Personal  Insurance. 

3.  Under  Property  Insurance  the  two  most  important 
divisions  are :  Fire  Insurance  and  Marine  Insurance. 

4.  Fire  Insurance  is  indemnity  against  loss  by  fire. 

5.  Marine  Insurance  is  indemnity  against  the  dangers 
of  navigation. 

NOTES. — 1.  Transit  Insurance  is  applied  to  risks  which  are  taken 
when  property  is  transferred  by  railroad,  or  by  railroad  and  water 
routes  combined. 

2.  There  are  several  minor  forms  of  property  insurance,  also, 
such  as  Live  Stock  Insurance,  Steam  Boiler  Insurance,  Plate  Glass  In- 
surance, etc.,  the  special  purposes  of  which  are  indicated  by  their 
names. 

6.  Personal  Insurance  is  of  three  kinds,   viz.:  Life  In- 
surance, Accident  Insurance,  and  Health  Insurance.     Personal 
insurance  will  be  discussed  in  another  chapter. 

7.  The  Insurer  or  Underwriter  is  the  party  or  company 
that  undertakes  to  pay  in  case  of  loss. 

8.  The  Risk  is  the  particular   danger  against  which  the 
insurer  undertakes. 

9.  The  Insured  is  the  party  protected  against  loss. 

10.  A    Contract  is  an  agreement  between    two  or  more 
competent   parties,  based  on  a  sufficient  consideration,  each 
promising   to   do   or  not  to  do  some  particular  thing  possi- 
ble to  be  done,  which  thing  is  not  enjoined  nor  prohibited 
by  law. 


INSURANCE.  229 

11.  The    Primary   Elements    of    a   contract    are:    the 
Parties,  the    Consideration,   the    Subject   Matter,    the   Consent 
of  the  Parties,  and  the   Time. 

12.  The  written  contract   between  the  two  parties  in  in- 
surance is  called  a  Policy. 

13.  The  Premium  is  the  sum  paid  for  insurance.     It  is 
a  certain  per  cent  of  the  amount  insured. 

14.  The  Rate  varies  with  the  nature  of  the  risk. 

15.  The  Amount  or  Valuation  is  the  sum  for  which  the 
premium  is  paid. 

NOTES. — 1.  Whoever  owns  or  has  an  interest  in  property,  may 
insure  it  to  the  full  amount  of  his  interest  or  liability. 

2.  Only  the  actual  loss  can  be  recovered  by  the  insured,  whether 
there  be  one  or  several  insurers. 

3.  Usually  property  is  insured  for  about  two  thirds  of  its  value. 

16.  Insurance  business  is  usually  transacted  by  incorpor- 
ated companies. 

17.  These  companies  are  either  joint-stock  companies,  or 
mutual  companies. 

KEMARKS. — 1.  In  joint-stock  companies  the  capital  is  owned  by 
individuals  who  are  the  stockholders.  They  share  the  profits  and 
losses. 

2.  In  mutual  companies  the  profits  and  losses  are  divided  among 
the  insured. 

275.  The  operations  in  insurance  are  included  under  the 
principles  of  Percentage. 

The  quantities  involved  are,  the  Amount  insured,  the 
Per  Cent  of  premium,  and  the  Premium. 

The  amount  corresponds  to  the  Base,  and  the  premium,  to 
the  Percentage. 

CASE    I. 

276.  Given  the  rate  of  insurance  and  the   amount 
insured,  to  find  the  premium. 

FORMULA. — Amount  Insured  X  Rate  —  Premium. 


230  RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  Insured  -f-  of  a  vessel  worth  $24000,  and  -f  of  its  cargo 
worth  $36000,  the  former  at  2%%,  the  latter  at  \\%\  what 
is  the  premium  ?  $607.50 

2.  Insured  a  house  for  $2500,  and  furniture  for  $600,  at 
T%/^:  what  is  the  premium?  $18.60 

3.  What  is  the  premium  on  a  cargo  of  railroad  iron  worth 
$28000,  at  lf%?  $490. 

4.  Insured  goods  invoiced  at  $32760,  for  three  months,  at 
T8o%:  what  is  the  premium?  $262.08 

5.  My  house  is  permanently  insured  for  $1800,  by  a  de- 
posit of  10  annual  premiums,  the  rate  per  year  being  |^: 
how  much  did  I  deposit,   and  if,  on  terminating  the  insur- 
ance, I  receive  my  deposit  less  5%,  how  much  do  I  get? 

$135  deposited;  $128.25  received. 

6.  A    shipment  of  pork,    costing    $1275,    is    insured    at 
-f^,  the   policy  costing  75   cents:  what  does  the  insurance 
cost?  $7.83 

7.  An  insurance  company   having  a  risk  of  $25000,   at 
T9o%,   re-insured   $10000,  at   |^,   with  another  office,  and 
$5000,  at  1^,  with  another:    how  much  premium  did   it 
clear  above  what  it  paid?  $95. 

CASE    II. 

277.     Given  the    amount   insured   and  premium,  to 
find  the  rate  of  insurance. 

Premium 

FORMULA. —  =  Rate. 

Amount  Insured 


EXAMPLES  FOR  PRACTICE. 

1.  Paid  $19.20   for  insuring  f  of  a  house,  worth 
what  was  the  rate? 


INSURANCE.  23: 

2.  Paid  $234,  including  cost  of  policy,  $1.50,  for  insuring 
a  cargo  worth  $18600:  what  was  the  rate?  \\% 

3.  Bought  books  in  England  for  $2468 ;  insured  them  fo 
the  voyage  for  $46.92,  including  the  cost  of  the  policy,  $2.50 
what  was  the  rate?  If  ^ 

4.  A    vessel    is    insured   for    $42000;    $18000    at    2%  ft 
$15000  at  3f%,  and  the  rest  at  4f%  :  what  is  the  rate  oj 
the  whole  $42000?  3f% 

5.  I   took    a    risk    of   $45000;    re-insured    at    the    sam 
rate,  $10000  each,  in  three  offices,  and  $5000  in  another 
my  share   of  the    premium    was    $262.50:    what    was    th 
rate?  2f% 

6.  I  took  a  risk  at  1^^  ;  re-insured  f  of  it  at  2^ ,  an< 
\  of  it  at  ^\%  •  what  rate  of  insurance  do  I  get  on  wha 
is  left?  ^% 

CASE    III. 

278.     Given  the  premium  and  rate  of  insurance,  t< 
find  the  amount  insured. 

Premium  T 

£  ORMULA. —  =  Amount  Insured. 

Rate. 


EXAMPLES  FOR  PRACTICE. 

1.  Paid  $118  for  insuring,  at  \%  :  what  was  the  amoun 
insured  ?  $14750 

2.  Paid  $411.37^  for  insuring  goods,  at  \\<fi)  :  what  wa 
their  value?  $27425 

3.  Paid  $42.30  for  insuring  f  of  my  house,  at  -$-§%  :  wha 
is  the  house  worth  ?  $7520 

4.  Took  a  risk   at   2^%  ;    re-insured    f    of   it   at  2|% 
my  share  of  the  premium  was  $197.13:  how  large  was  th 
risk?  $26284 

5.  Took  a  risk  at  lf%;  re-insured  half  of  it  at  the  sam< 
rate,  and  ^  of  it  at  1^-^  ;  my  share  of  the   premium  wa 
$58.11 :  how  large  was  the  risk?  $19370 


232  RA  Y^S  HIGHER  ARITHMETIC. 

6.  Took  a  risk  at  2%;  re-insured  $10000  of  it  at  1\%, 
and  $8000  at  \\%\  my  share  of  the  premium  was  $207.50: 
what  sum  was  insured  ?  $28000. 

7.  The  Mutual  Fire  Insurance  Company  insured  a  build- 
ing and  its    stock  for  |-  of   its  value,  charging  lf^.     The 
Union  Insurance  Company  relieved  them  of  \  of  the  risk, 
at  \\%.     The  building  and  stock  being   destroyed  by  fire, 
the  Union   lost    forty-nine    thousand    dollars    less    than  the 
Mutual:    what    amount    of  money   did   the  owners   of  the 
building  and  stock  lose  ?  $51 750. 


VII.    TAXES. 
DEFINITIONS. 

279.  1.  A  Tax  is  a  sum  of  money  levied  on  persons,  or 
on  persons  and  property,  for  public  use. 

2.  Taxes  in  this  country  are:   (1.)  State  and  Local  Taxes; 
(2.)   Taxes  for  the  National  Government. 

3.  Taxes  are  further  classified  as  Direct  and  Indirect. 

4.  A  Direct  Tax  is  one  which  is  levied  on  the  person  or 
property  of  the  individual. 

5.  An   Indirect  Tax  is  a  tax  levied  on  articles  of  con- 
sumption, for  which  each  person  pays  in  proportion  to  the 
quantity  or  number  of  such  articles  consumed. 

6.  A  Poll-Tax,  or  Capitation  Tax,  is  a  direct  tax  levied 
on  each  male  citizen  liable  to  taxation. 

7.  A  Property  Tax  is  a  direct  tax  levied  on  property. 

NOTE. — In  legal  works  property  is  treated  of  under  two  heads  ; 
viz  ,  Real  Property,  or  Real  Estate,  including  houses  and  lands  ;  and 
Personal  Property,  including  money,  bonds,  cattle,  horses,  furniture, — 
in  short,  all  kinds  of  movable  property. 


TAXES.  233 

8.  State  and  Local  Taxes  are  generally  direct,  while  the 
United  States  Taxes  are  indirect. 

9.  An  Assessor  is  a  public  officer  elected  or  appointed  to 
prepare  the   Assessment  Roll. 

10.  An  Assessment   Roll  is  a  list  of  the  names  of  the 
taxable  inhabitants   living  in  the  district   assessed,  and  the 
valuation  of  each  one's  property. 

11.  The  Collector  is  the  public  officer  who  receives  the 
taxes. 

NOTE. — In  some  states  all  the  taxes  are  collected  by  the  counties ; 
in  others,  the  towns  collect;  while  in  others  the  collections  are 
made  by  separate  collectors.  Generally  a  number  of  different  taxes 
are  aggregated, — such  as  state,  county,  road,  school,  etc. 

280.  The  quantities   involved   in  problems  under  tax- 
ation are,  the  Assessed    Value  of  the  property,  the  Rate  of 
Taxation,  the  Tax,  and  the  Amount  Left  after  taxation. 

They  require  the  application  of  the  four  cases  of  Percent- 
age, the  assessed  value  corresponding  to  the  Base;  the  tax, 
to  the  Percentage;  and  the  amount  left  after  taxation,  to 
the  Difference. 

CASE    I. 

281.  Given  the    taxable  property  and  the   rate,  to 
find  the  property- tax. 

FORMULA. — Taxable  Property  X  Rate  =  Amount  of  Tax. 

NOTE. — If  there  be  a  poll-tax,  the  sum  produced  by  it  should  be 
added  to  the  property-tax,  to  give  the  whole  tax. 


EXAMPLES  FOR  PRACTICE. 

1.  The  taxable  property  of  a  county  is  $486250,  and  the 
rate  of  taxation  is  78  ct.  on  $100 ;  that  is,  yVo^  :  what  is 
the  tax  to  be  raised  ?  $3792.75 

H.  A.  20. 


234 


RAY'S  HIGHER  ARITHMETIC. 


REMARK. — The  rate  of  taxation  being  usually  small,  is  expressed 
most  conveniently  as  so  many  cents  on  $100,  or  as  so  many  mills 

on  $1. 

2.  A's  property  is  assessed  at  $3800 ;  the  rate  of  taxation 
is  96  ct.  on  $100  (T9o^0  •  what  is  his  whole  tax,  if  he 
pays  a  poll-tax  of  $1?  $37.48 

REMARKS. — 1.  In  making  out  bills  for  taxes,  a  table  is  used,  con- 
taining the  units,  tens,  hundreds,  thousands,  etc.,  of  property,  with 
the  corresponding  tax  opposite  each. 

2.  To  find  the  tax  on  any  sum  by  the  table,  take  out  the  tax  on 
each  figure  of  the  sum,  and  add  the  results.  In  this  table,  the  rate 
is  I\<f0,  or  125  ct.  on  $100. 


TAX  TABLE. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1 

.0125 

$10 

.125 

$100 

$1.25 

$1000 

$12.50 

$10000 

$125 

2 

.025 

20 

.25 

200 

2.50 

2000 

25. 

20000 

250 

3 

.0375 

30 

.375 

300 

3.75 

3000 

37.50 

30000 

375 

4 

.05 

40 

.50 

400 

5. 

4000 

50. 

40000 

500 

5 

.0625 

50 

.625 

500 

6.25 

5000 

62.50 

50000 

625 

6 

.075 

60 

.75 

600 

7.50 

6000 

75. 

60000 

750 

7 

.0875 

70 

.875 

700 

8.75 

7000 

87.50 

70000 

875 

8 

.10 

80 

1. 

800 

10. 

8000 

100. 

80000 

1000 

9 

.1125 

90 

1.125 

900 

11.25 

9000 

112.50 

90000 

1125 

3.  What  will  be  the  tax  by  the  table,  on  property  assessed 
at  $25349  ? 

SOLUTION.— The  tax  for  $20000  is  $250;  for  $5000  is  62.50;  for 
$300  is  $3.75;  for  $40  is  .50;  for  $9  is  .1125;  which,  added,  give 
$316.86,  the  tax  on  $25349. 

4.  Find  the  tax  for  $6815.30  $85.19 

5.  What  is  the   tax  on   property  assessed  at  $10424.50, 
and  two  polls,  at  $1.50  each?  $133.31 

6.  A's  property  is  assessed  at  $251350,  and  B's  at%  $25135. 
What  is  the  difference  in  their  taxes?  $2827.69  — 


TAXES.  235 


CASE   II. 

282.     Given  the   taxable    property  and   the   tax,  to 
find  the  rate. 

Tax 

FOKMULA.—  — -  —  =  Bate. 

Taxable  Property 


EXAMPLES  FOR  PRACTICE. 

1.  Property  assessed  at  $2604,  pays  $19.53  tax:  what  is 
the  rate  of  taxation  ?  \%  =  75  ct.  on  $100. 

2.  The   taxable    property   in    a    town   of    1742    polls,   is 
$6814320;  a  tax  of  $66913  is  proposed:    if  a  poll-tax   of 
$1.25  is  levied,  what  should  be  the  rate  of  taxation? 

T9-o5o%^95ct.  on  $100. 

3.  An  estate  of  $350000  pays  a  tax  of  $5670 :  what  is  the 
rate  of  taxation  ?  \\\%  =  $1.62  on  $100. 

4.  A's    tax    is    $50.46 ;    he    pays    a    poll-tax    of    $1.50, 
and  owns  $8704  taxable  property :  what  is  the  rate  of  tax- 
ation? -£sf0  ===  561  ct.  on  $100. 

CASE    III. 

283.     Given  the  tax  and  the  rate,  to  find  the  assessed 
value  of  the  property. 

Tax 

FORMULA. :—  =  Taxable  Property. 

Rate 

NOTE. — If  any  part  of  the  tax  arises  from  polls,  it  should  be  first 
deducted  from  the  given  tax. 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  assessed  value  of  property  taxed  $66.96, 
at  If  %  ?  $3720. 


236  RAY'S  HIGHER  ARITHMETIC. 

2.  A  corporation  pays  $564.42  tax,  at  the  rate  of  T40 

or  46  ct.  on  $100  :  find  its  capital.  $122700. 

3.  A  is  taxed  $71.61  more  than  B;  the  rate  is  l^g- %,  or 
$1.32  on  $100:  how  much  is  A  assessed  more  than  B? 

$5425. 

4.  A  tax  of  $4000  is  raised  in  a  town  containing  1024 
polls,  by  a  poll-tax  of  $1,  and  a  property-tax  of  T2^\^  (24 
ct.  on  $100)  :  what  is  the  value  of  the  taxable  property  in 
it  ?  $1240000. 

5.  A's  income  is  16%  of  his  capital;  he  is  taxed  %\%  of 
his  income, %and  pays  $26.04:  what  is  his  capital?       $6510. 

CASK    IV. 

284.  Given  the  amount  left  after  payment  of  tax 
and  the  rate,  to  find  the  assessed  value  of  the  prop- 
erty. 

Amount  Left  after  Payment       rrt 

FORMULA. J- — J—  — =  Taxable  Property. 

1  —  Rate  o/c. 


EXAMPLES  FOR  PRACTICE. 

1.  A  pays  a  tax  of  1-fafi  ($1.35  on  $100)  on  his  capital, 
and    has  left    $125127.66:    what   was  his   capital,  and  his 
tax?  Capital,  $126840;  tax,  $1712.34 

2.  Sold  a  lot  for  $7599,  which  covered  its  cost  and  2^ 
beside,  paid  for  tax :  what  was  the  cost  ?  $7450. 


VIII.    UNITED  STATES  EEVENUE. 
DEFINITIONS. 

285.  1.  The  Revenue  of  the  United  States  arises  from 
the  Internal  Revenue*  from  the  Qustoms,  and  from  Sales  of 
Public  Lands. 


UNITED  STATES  REVENUE.  237 

2.  The  Internal  Revenue  is  derived  from  taxes  on  spirits, 
tobacco,  fermented  liquors,   banks,    and   from    the    sale   of 
stamps,  etc. 

3.  The  Customs  or  Duties   are   taxes  imposed  by  Gov- 
ernment on   imported   goods.      They   are  of  two  kinds, — 
ad  valorem  and  specific. 

4.  Ad  Valorem  Duties  are  levied  at  a  certain  per  cent 
on  the  cost  of  the  goods  as  shown  by  the  invoice. 

5.  Specific    Duties   are    certain  sums  collected  on  each 
gallon,  bushel,  yard,  ton,  or  pound,  whatever  may  be  the 
cost  of  the  article. 

NOTE. — Problems  involving  specific  duty  only  are  not  solved,  of 
course,  by  the  principles  of  Percentage. 

6.  An  Invoice  is  a  detailed  statement  of  the  quantities 
and  prices  of  goods  purchased. 

7.  A  Tariff  is  a  schedule  of  the  rates  of  duties,  as  fixed 
by  law. 

NOTES. — 1.  The  collection  of  duties  is  made  at  the  custom-houses 
established  at  ports  of  entry  and  ports  of  delivery.  The  principal 
custom-house  officers  are  collectors,  naval  officers,  surveyors,  and 
appraisers. 

2.  Tare  is  an  allowance  for  the  weight  of  whatever  contains  the 
goods.     Duty  is  collected  only  on  the   quantities   passed  through 
the  custom-house.     The  ton,  for  custom-house  purposes,  consists  of 
20  cwt.,  of  112  Ib.  each. 

3.  On  some  classes  of  goods,  both  specific  and  ad  valorem  duties 
are  collected.     The  cost  price,  if  given  in  foreign  money,  must  be 
changed  to  United  States  currency. 

286.  Problems  in  United  States  Customs,  where  the 
duty  is  wholly  or  in  part  ad  valorem,  are  solved  by  the 
principles  of  Percentage. 

The  quantities  involved  are :  the  Invoice  Price  corre- 
sponding to  the  Base;  the  Duty,  corresponding  to  the  Per- 
centage; and  the  Total  Cost  of  the  importation,  corresponding 
to  the  Amount. 


238  BAY'S  HIGHER  ARITHMETIC. 


CASE    I. 


287.     Given  the  invoice  price  and  the   rate,  to  find 
the  duty. 

FORMULA. —  Invoice  Price  X  -Rate  =  Duty. 


EXAMPLES  FOR  PRACTICE. 

1.  Import   24  trunks,  at  $5.65  each,  and  3  doz.  leather 
satchels,  .at  $2.25  each;  the  rate  is  35^  ad  valorem:  what 
is  the  duty?  $75.81 

2.  What  is  the  duty  on  45  casks  of  wine,  of  36  gal.  each, 
invoiced  at  $1.25  a  gal.,  at  40  ct.  a  gal.  specific  duty?    $648. 

3.  There  is  a  specific  duty  of  $3  per  gallon,  and  an  ad 
valorem  duty  of  50%  on  cologne-water:    what  is  the  total 
amount  of  duty  paid  on  25  gallons,  invoiced  at  $16.50  per 
gallon?  $281.25 

4.  What  is  the  total  duty  on  36  boxes  of  sugar,  each 
weighing  6  cwt.  2  qr.  18  lb.,  invoiced  at  2-|-  ct.  per  lb.,  the 
specific  duty  being  2  ct.  per  lb.,  and  the  ad  valorem  duty 

$704.97 

5.  What  is  the  duty  on  575  yards  of  broadcloth,  weighing 
1154  lb.,  invoiced  at  $2.56  per  yd.,  the  specific  duty  being 
50  ct.  per  lb.,  and  the  ad  valorem  duty  35^  ?    If  the  freight, 
charges,  and  losses   amount  to  $160.80,  how  much  a  yard 
must  I  charge  to  gain  15%  ? 

Duty,  $1092.20;  price  per  yard,  $5.45 

6.  A   merchant  imported  a  ton  of  manilla,  invoiced   at 
5  ct.  per  lb.;  he  paid  a  specific  duty  per  ton,  which,  on  this 
shipment,  was  equivalent  to  an  ad  valorem  duty  of  22^^  : 
what  is  the  specific  duty  ?  $25  per  ton. 

7.  Received  a  shipment  of  3724  lb.  of  wool,  invoiced  at 
23  ct.  per  lb.;  the  duty  is  10  ct.  per  lb.,  and  11^  ad  valorem, 
less  lO^g:  what  is  the  total  amount  of  duty?  $419.96 


UNITED  STATES  REVENUE.  239 

8.  A  dry-goods  merchant  imports  1120  yards  of  dress 
goods,  1|  yd.  wide,  invoiced  at  23  ct.  a  sq.  yd.;  there  is  a 
specific  duty  of  8  ct.  per  sq.  yd.,  and  an  ad  valorem  duty 
of  40^:  what  must  he  charge  per  yard,  cloth  measure,  to 
clear  25^  on  the  whole?  $.62|f  per  yd. 

CASE    II. 

288.  Given  the  invoice  price  and  the  duty,  to  find 
the  rate. 

Duty 

FORMULA. —  — -       *   .-  •  =  Rate. 
Invoice  Price 


EXAMPLES  FOR  PRACTICE. 

1.  If  goods  invoiced  at  $3684.50  pay  a  duty  of  $1473.80, 
what  is  the  rate  of  duty?  40%. 

2.  If  laces    invoiced    at    $7618.75,    cost,    when    landed, 
$10285.311,  what  is  the  rate  of  duty?  35^. 

3.  Forty  hhd.  (63  gal.  each)  of  molasses,  invoiced  at  52 
ct.  a  gallon,  pay  $453.60  duty.     The  specific  duty  is  5  ct.  a 
gallon:  what  is  the  additional  ad  valorem  duty?  25%. 

CASE   III. 

289.     Given  the    duty  and   the  rate,  to  find  the  in- 
yoice  price. 

FORMULA. -~~  —  Invoice  Price. 

Rate 


EXAMPLES  FOR  PRACTICE. 

1.  Paid  $575.80  duty  on  watches,  at  25%:  at  what  were 
they  invoiced,  and  what  did  they  cost  me  in  store? 

Invoiced,  $2303.20;  cost,  $2879. 

2.  The    duty  on  1800  yards  of  silk  was  $2970,    at  60% 


240  RA  Y>  S  HIGHER  ARITHMETIC. 

ad  valorem :  what  was  the  invoice  price  per  yard,  and  what 
must  I  charge  per  yard  to  clear  20%  ? 

Invoiced  at  $2.75;  sell  at  $5.28  per  yd. 
3.  The  duty  on  15  gross,  qt.  bottles  of  porter,  at  a  tax  of 
35  ct.  a  gallon,  was  $151.20:  if  this  were  equivalent  to  an 
ad  valorem  duty  of  %§-f§%  on  the  entire  purchase,  how 
many  bottles  were  allowed  for  a  gallon,  and  at  how  much 
per  bottle  must  the  whole  be  sold  to  clear  20%  ? 

5  bottles  to  gal. ;  50  ct.  per  bottle. 

CASE    IV. 

290.     Given  the  entire  cost  and  the  rate,  to  find  the 
invoice  price. 

Whole  Cost 

FORMULA. —  =  Invoice  Price. 

1  +  Rate 


EXAMPLES  FOR  PRACTICE. 

1.  1000  boxes  (100  each)  pf  cigars,  weighing  1200  lb.,  net, 
cost  in  store  $13675.     There  is  a  specific  duty  of  $2.50  per 
lb.,  an  ad  valorem   duty  of  25%,  and   an  internal  revenue 
tax  of  60  ct.   a  box :   freight  and  charges  amount  to  $75  ; 
find  the  invoice  price  per  thousand  cigars.  $80. 

2.  Supposing  No.  1  pig-iron,  American  manufacture,  to  be 
of  equal  quality  with   Scotch  pig-iron :    at  what  price  must 
the  latter  be  invoiced,  to  compete  in  our  markets,  if  Amer- 
ican iron  sells  for  $45  a  ton ;  freight  and  charges  amounting 
to  $10  a  ton,  and  the  specific  duty  being  equivalent,  in  this 
instance,  to  an  ad  valorem  duty  of  25%  ?  $28  a  ton. 

3.  A  marble-cutter  imports  a  block  of  marble  6  ft.  6  in. 
long,  3  ft.  wide,  2  ft.  9f  in.  thick  ;    the  whole  cost  to  him 
being  $130 ;  he  pays  a  specific  duty  of  50  ct.  per  cu.  ft.  and 
an  ad  valorem   duty  of   20^  :    freight    and   charges    being 

3.80,  what  was  the  invoice  price  per  cu.  ft.?  $1.25 


TOPICAL    OUTLINE. 


241 


Topical   Outline. 
APPLICATIONS  OF  PERCENTAGE. 

(Without  Time.) 


1.  Profit  and  Loss.... 


2.  Stocks  and  Bonds.. 


3.  Premium  and  Discount 


4.  Commission  and  Brokerage. 


5.  Stock  Investments.. 


6.  Insurance., 


7.  Taxes.. 


8.  United  States  Revenue 


1.  Definitions :— Cost,    Selling    Price,   Profit, 

Loss. 

2.  Four  Cases. 

1.  Definitions: — Company,  Corporation,  Char- 

ter, Stock,  Shares,  Scrip,  Bond,  Assess- 
ment, Dividend. 

2.  Four  Cases. 

1.  Definitions :— Drafts,    Par   Value,    Market 

Value,  Discount,  Premium,  Rate. 

2.  Four  Cases. 

l.  Definitions:— Commission-Merchant,  Prin- 
cipal,   Commission,    Consignment,    Con- 
signor,   Consignee,    Correspondent,    Net 
Proceeds,  Guaranty,  Broker,  Brokerage. 
(^  2.  Four  Cases. 

I    1.  Definitions: — Stock  Exchange,  Government 

(Bonds,  State  Bonds,  etc. 
2.  Notation. 
3.  Five  Cases. 

1.  Definitions: — Fire  Insurance,  Marine   In- 

surance, Personal  Insurance,  Insurer, 
Risk,  Insured,  Contract,  Primary  Ele- 
ments, Policy,  Premium,  Rate,  Amount, 
Joint-Stock  and  Mutual  Companies. 

2.  Three  Cases. 

1.  Definitions :— Tax,  State  and  Government, 

Direct  and  Indirect,  Poll-tax,  Property- 
tax,  Assessor,  Assessment  Roll,  Collector. 

2.  Four  Cases. 

1.  Definitions:— Internal  Revenue,  Customs, 

Ad  Valorem  Duties,  Specific  Duties,  In- 
voice, Tariff,  Tare. 

2.  Four  Cases. 


H.  A.  21. 


XVI.   PEKCEOTAGE  -APPLICATIONS. 

I.   INTEKEST. 
DEFINITIONS. 

291.     1.  Interest  is  money  charged  for  the  use  of  money. 

NOTE. — The  profits  accruing  at  regular  periods  on  permanent 
investments,  such  as  dividends  or  rents,  are  called  interest,  since 
they  are  the  increase  of  capital,  unaided  by  labor. 

2.  The  Principal  is  the  sum  of  money  on  which  interest 
is  charged. 

NOTE. — The  principal  is  either  a  sum  loaned ;  money  invested  to 
secure  an  income ;  or  a  debt,  which  not  being  paid  when  due,  is 
allowed  by  agreement  or  by  law  to  draw  interest. 

3.  The  Kate  of  Interest  is  the  number  of  per  cent  the 
yearly  interest  is  of  the  principal. 

4.  The  Amount  is  the  sum  of  the  principal  and  interest. 

5.  Interest  is  Payable  at  regular   intervals,  yearly,  half- 
yearly,  or  quarterly,  as  may  be  agreed:  if  there  is  no  agree- 
ment, it  is  understood  to  be  yearly. 

NOTE. — If  interest  is  payable  half-yearly,  or  quarterly,  the  rate 
is  still  the  rate  per  annum,  or  rate  per  year.  In  short  loans,  the  rate 
per  month  is  generally  given ;  but  the  rate  per  year,  being  12  times 
the  rate  per  month,  is  easily  found ;  thus,  2^>  a  month  =  24^  a 
year. 

6.  The  Legal  Kate  is  the   highest    rate  allowed   by  the 
law. 

7.  If  interest  be  charged  at   a  rate  higher  than  the  law 
allows,  it   is  called  Usury;   and,  in  some  states,  the  person 
offending  is  subject  to  a  penalty. 

(242) 


INTEREST. 


243 


REMARK. — The  per  cent  of  interest  that  is  legal  in  the  different 
states  and  territories,  is  exhibited  in  the  following  table. 

TABLE. 


NAME   OF  STATE. 

RATE. 

NAME   OF  STATE. 

RATE. 

Alabama   

8% 
10* 

6% 

10% 
6% 
10% 
6% 
7% 
6% 
6% 
8% 
7% 
10% 
6% 
6% 
6% 
7% 
6% 
5% 
6% 
6% 
6% 
7% 
7% 
6% 

8% 
Any. 

Wfo 
Any. 

6% 
Any. 

6% 
12% 
6% 
10% 
Any. 
Any. 
24% 
8% 
8% 
10% 
12% 
10% 
8% 
Any. 
6% 
6% 
10% 
10% 
10% 

Missouri  • 

6% 
10% 
10% 
10% 
6# 
6% 
6% 
6% 
6% 
6% 
10% 
6% 
6% 
7% 
6% 
8% 
6% 
10% 
6% 
6% 

10% 
6% 
7% 
12% 

10% 
Any. 

12% 
Any. 
6% 
6% 
12% 
6% 
8% 
8% 
12% 
6% 
Any. 
Any. 
6% 
12% 
6% 
Any. 
6% 
8% 

Any. 

6% 
10% 
Any. 

_ 

Montana  

\rkansas  

Nebraska 

California  

Nevada 

Canada  

New  Hampshire... 

N"ew  Jersev 

Colorado  

Connecticut  

New  M^exico««« 

Dakota  

New  ^Tork 

Delaware            • 

North  Carolina  
Ohio  

District  Columbia. 
Florida  

Oreffon  

Oeorsria  

Pennsylvania 

Idaho  

Ehode  Island  
South  Carolina  
Tennessee 

Illinois  oo.. 

Indiana    

Iowa  

Texas            

United  States  

Utah  

Vermont  

Maine  » 

Virginia  

Washington     Ter- 
ritory   

Massachusetts  
Michigan  

West  Virginia  
AVisconsin  

Minnesota     

Mississippi  

\Vvomm01 

i 

NOTE. — When  the  per  cent  of  interest  is  not  mentioned  in  the 
note  or  contract,  the  first  column  gives  the  per  cent  that  may  be 
collected  by  law.  If  stipulated  in  the  note,  a  per  cent  of  interest  as 
high  as  that  in  the  second  column  may  be  collected. 

8.  Interest  is  either  Simple  or  Compound. 

9.  Simple  Interest   is  interest  which,  even    if   not  paid 


244  RAY'S  HIGHER  ARITHMETIC. 

when  due,  is  not  convertible  into  principal,  and  therefore 
can  not  accumulate  in  the  hands  of  the  debtor  by  drawing 
interest  itself,  however  long  it  may  be  retained. 

NOTE. —  Compound  Interest  is  interest  which,  not  being  paid  when 
due,  is  convertible  into  principal,  and  from  that  time  draws  interest 
itself  and  accumulates  in  the  hands  of  the  debtor,  according  to  the 
time  it  is  retained.  It  will  be  treated  of  in  a  separate  chapter. 

292.  Simple    Interest   differs   from    the    applications    of 
Percentage  in  Chapter  XV,  by  taking   time  as  an  element 
in  the  calculation,  which  they  do  not. 

293.  The  five  quantities  embraced   in   questions  of  in- 
terest are:    the  Principal,  the  Interest,  the  Rate,  the  Time, 
and  the  Amount, — or  the  sum  of  the  principal  and  interest. 
Any  three  of  these  being  given,  the  others  may  be  found. 
They  give  rise  to    five  cases. 

The  principal  corresponds  to  the  Rase,  and  the  interest 
to  the  Percentage. 

NOTATION. 

294.  The    following   notation    can    be    adopted    to    ad- 
vantage  in  the  formulas :  « 

Principal  =  P. 
Eate         ==  K. 
Interest    =  I. 
Amount  =  A. 
Time         =  T. 


CASE    I. 

295.     Given  the   principal,  the    rate,  and   the  time, 
to  find   the   interest  and  the  amount. 

PRINCIPLE. — The  interest  is  equal  to  the  continued  product 
of  the  principal,  rate,  and  time. 


INTEREST.  245 

f  P  V  E)  \^  T  =^:  I 

FORMULAS. —     pi       -  A 

NOTE. — The  time  is  expressed  in  years  or  parts  of  years,  or  both. 

COMMON  METHOD. 

PROBLEM. — What  is  the  interest  of  $320  for  3  yr.  5  mo. 
18  da.,  at  4%  ? 

OPERATION. 

r  =  $44.37i,  Int. 


SOLUTION.— The  interest  of  $320  for  1  year,  at  4^,  is  $12.80 ;  and 
for  3^  years,  it  is  3^  times  as  much  as  it  is  for  1  year,  or  $12.80  X 
3A  =  $44.37J. 

REMARK. — Unless  otherwise  specified,  30  days  are  considered  to 
make  a  month;  hence,  5  mo.  18  da.  =  xV  °^  a  ye 


General  Rule. — 1.  Multiply  the  principal  by  the  rate,  and 

that  product  by  the  time  expressed  in  years;  the  product  is  the 
interest. 

2.  Add  Hie  principal  and  interest,  to  find  the  amount. 

EXAMPLES  FOR  PRACTICE. 

Find  the  simple  interest  of: 

1.  $178.63  for  2  yr.  5  mo.  26  da.,  at  lfc.  $31.12 

2.  86084.25  for  1  yr.  3  mo.,  at  4^.    '  $342.24 

3.  $64.30  for  1  yr.  10  mo.  14  da.,  at  9%.  $10.83 

4.  $1052.80   for   28  da.,  at  10%.  $8.19 

5.  $419.10  for  8  mo.  16  da.,  at  6%.  $17.88 

6.  $1461.85  for  6  yr.  7  mo.  4  da.,  at  10%.  $964.01 

7.  $2601.50  for  72  da.,  at  1\%.  $39.02 

8.  $8722.43  for  5|  yr.,  at  6%.  $2878.40 

9.  $326.50  for  1  mo.  8  da.,  at  8%.  $2.76 

10.  $1106.70  for  4  yr.  1  mo.  1  da.,  at  &%.  $271.33 

11.  $10000  for  1  da.,  at  Q%.  $1.67 


246 


RAY'S  HIGHER   ARITHMETIC. 


METHOD   BY   ALIQUOT  PAKTS. 

296.     Many  persons    prefer    computing    interest    by   the 
method  of  Aliquot  Parts.     The  following  illustrates  it : 

PROBLEM. — What  is   the   simple   interest  of  $354.80  for 
3  yr.  7  mo.  19  da.,  at  6%? 

SOLUTION. — The  yearly  in- 
terest, being  6$>  of  the  prin- 
cipal, is  found,  by  Case  I  of 
Percentage;  the  interest  for  3 
yr.  7  mo.  19  da.  is  then  ob- 
tained by  aliquot  parts;  each 
item  of  interest  is  carried  no 
lower  than  mills,  the  next 
figure  being  neglected  if  less 
than  5 ;  but  if  5  or  over,  it  is 
counted  1  mill. 


OPERATION. 

$354.80 

.06 

l  yr.  = 

21.2880 

3  yr.   =  3 

63.864 

6  mo.  =  ^ 

10.644 

1  mo.  =  1 

1.774 

1  8  da.  ==  A 

1.064 

1  da.  =  /o 

.059 

$77.41,  Ans. 


SIX  PEE   CENT  METHODS. 
FIRST  METHOD. 

297.     The  six  per  cent  method  possesses  many  advantages, 
and  is  readily  and  easily  applied,  as  the  following  will  show  : 

At  6$)  per  annum,  the  interest  on  $1 

for  1  year  is    6    ct.,  or  .06       of  the  principal. 

"    2  mo.  "    1     "     "    .01       "      "  " 

tt    i    tt  u    i     tt     «    t0Q5     u      «  <c 

"    6  da.  =  £  mo.   "   TV    "     "    .001      "      "  " 


PROBLEM.  —  What  is  the  interest  of  $560  for  3  yr.  8  mo. 
12  da.,  at  6%? 

SOLUTION.  —  The  interest  on  OPERATION. 

$1  for  1  yr.,  at  6^,  is  6  ct.,  and  $560X-222  =  $124.32,  Ans. 
for  3  yr.  is  18  ct.;  the  interest 

on  $1  for  8  mo.  is  4  ct.,  and  the  interest  for  12  da.  is  2  mills  ;  hence, 
the  interest  for  the  entire  time  is  $.222  :  therefore,  the  interest  on 
$560  for  3  yr.  8  mo.  12  da.  is  equal  to  $.222  X  560  =  $124.32 


INTEREST.  247 

Six  Per  Cent  Rule. — 1.  Take  six  cents  for  every  year,  J  a 
cent  for  every  month,  and  ^  of  a  mill  for  every  day ;  their  sum 
is  the  interest  on  $1,  at  6%,  for  the  given  time. 

2.  Multiply  the  interest  on  $1  for  the  given  time  by  the  prin- 
cipal; the  product  is  the  interest  required. 

REMARKS. — 1.  To  find  the  interest  at  any  other  rate  than  6^,  in- 
crease or  decrease  the  interest  at  6^>  by  such  a  part  of  it  as  the  given 
rate  is  greater  or  less  than  6^. 

2.  After  finding  the  interest  at  6^,  observe  that  the  interest  at 
5^  =  interest  at  Qfc  —  |  of  itself.  And  the  interest 


at  4J^£  =  int.  at  6f0  —  J  of  it. 
at  4  fc  =  int.  at  6$  —  J  of  it. 
at  3  cf0  =  J  int.  at  6^>. 
at  2  ff0  =  \  int.  at  6^>. 
at  \\<j0-=.\  int.  at  6<fi. 


at  1  f0  =  i  int.  at  6^0. 
at  7  ^  =  int.  at  Qf0  +  i  of  it. 
at  7%f0  =      «      6/0  +  i  of  it. 
at  8  cj0  =      "      6fo  +  3  of  it. 
at  9  ft  =       "      6^>  +  J  of  it. 


at  12^,  18^,,  24^,  =  2,  3,  4  times  interest  at  6^,. 

at  5^0,  10^,,  15^,  20/0  =TV,  i,  J,  J  interest  at  6^0,  after  moving 
the  point  one  figure  to  the  right. 

3.  Or,  having  the  interest  at  6^,,  multiply  by  the  given  rate  and 
divide  bj  6. 

SECOND  METHOD. 

PROBLEM.  —  What  is  the  simple  interest  of  $354.80  for  3 
yr.  7  mo.  19  da.,  at 


SOLUTION.  —  The  interest  of  any  OPERATION. 

sum  ($354.80),  at  6^0,  equals  the  ($354.80^-2)X.436J  = 
interest  of  half  that  sum  ($177.40),  $77.40553,  Ans. 

at  12^,.      But  12^  a  year  equals 

l<fa  a  month,  and  for  3  yr.  7  mo.,  or  43  mo.,  the  rate  is  43^,,  and  for 
19  da.,  which  is  -J-g  of  a  month,  the  rate  is  Jfj^;  hence,  the  rate  for 
the  whole  time  is  43|f  ^,  and  43J§^>  of  the  principal  will  be  the 
interest.  To  get  43J§^,  multiply  by  43-J-j}  hundredths  ==  .43|f  = 
.436J. 

KEMARK.  —  In  the  multiplier  .436J,  the  hundredths  (43)  are  the 
number  of  months,  and  the  thousandths  (6J)  are  £  of  the  days 
(19  da.),  in  the  given  time,  3  yr.  7  mo.  19  da. 


248 


RAY'S  HIGHER   ARITHMETIC. 


Rule. — Reduce  the  years,  if  any,  to  months,  and  write  the 
whole  number  of  months  as  decimal  hundredths;  after  which, 
place  J  of  the  days,  if  any,  as  thousandths;  multiply  half  the 
principal  by  this  number,  the  product  is  the  interest. 

NOTE. — In  applying  the  rule,  when  the  number  of  days  is  1  or  2, 
place  a  cipher  to  the  right  of  the  months,  and  write  the  J  or  | ; 
otherwise,  they  will  not  stand  in  the  thousandths'  place :  thus,  if 
the  time  is  1  yr.  4  mo.  1  da.,  the  multiplier  is  .160J 

REMARK. — Exact  or  Accurate  Interest  requires  that  the  common 
year  should  be  365,  and  leap  year  366  days ;  hence,  exact  interest  is 
7*3  less  for  common  years,  and  -^  less  for  leap  years  than  the  ordi- 
nary interest  for  360  days. 


EXAMPLES  FOR  PRACTICE. 

Find  the  simple  interest  of: 

1.  $1532.45  for  9  yr.  2  mo.  7  da.,  at  12%.  $1689.27 

2.  $78084.50  for  2  yr.  4  ruo.  29  da.,  at  18^.  $33927.72 

3.  $512.60  for  8  mo.  18  da.,  at  1%.  $25.72 

4.  $1363.20  for  39  da.,  at  \\%  a  month.  $22.15 

5.  $402.50  for  100  da.,  at  2^  a  month.  $26.83 

6.  $6919.32  for  7  yr.  6  mo.,  at  6%.  $3113.69 

7.  $990.73  for  9  mo.  19  da.,  at  1%.  $55.67 

8.  $4642.68  for  5  mo.  17  da.,  at  15%.  $323.05 

9.  $13024  for  9  mo.  13  da.,  at  Wfi.  $1023.83 

10.  $615.38  for  4  yr.  11  mo.  6  da.,  at  20%.  $607.17 

11.  $2066.19  for  3  yr.  6  mo.  2  da.,  at  30^.  $2172.94 

12.  $92.55  for  3  mo.  22  da.,  at  5%.  $1.44 

Find  the  amount  of: 

13.  $757.35  for  117  da.,  at  \\<fc  a  month.  $801.65 

14.  $1883  for  1  yr.  4  mo.  21  da.,  at  6%.  $2040.23 

15.  $262.70  for  53  da.,  at  1%  a  month  $267.34 

16.  $584.48  for  133  da.,  at  1\%.  $600.67 

17.  $392.28  for  71  da.,  at  Z\%  a  month.  $415.49 


INTEREST.  249 

18.  Find   the  interest   of  $7302.85  for  365  da.,   at  6%, 
counting  360  da.  to  a  year.  $444.26 

19.  If  I  borrow  $1000000  in  New  York,  at  1%,  and  lend 
it  at  lfc  in  Ohio,  what  do  I  gain  in  180  da.?  $479.45 

NOTE. — In  New  York  365  days  are  counted  a  year  instead  of  360. 

20.  Find  the  interest  of  $5064.30  for  7  mo.  12  da.,  at  1%, 
in  New  York.  $218.45 

21.  If  I  borrow  $12500  at  6%,  and  lend  it  at  10^,  what 
do  I  gain  in  3  yr.  4  mo.  4  da.?  $1672.22 

22.  If  $4603.15  is  loaned  July  17,  1881,  at  7%,  what  is 
due  March  8,  1883?  $5130.34 

NOTE. — When  the  time  is  to  be  found  by  subtraction,  and  the  days 
in  the  subtrahend  exceed  those  in  the  minuend,  disregard  days  in 
subtracting,  and  take  the  months  in  the  minuend  one  less  than  the 
actual  count.  Having  found  the  years  and  months  by  subtraction, 
find  the  days  by  actual  count,  beginning  in  the  month  preceding  the 
later  of  the  two  dates.  Thus,  in  the  above  example,  we  find  1  yr. 

7  mo.  (not  8),  and  count  from  February  17th  to  March  8th  19  da. 

23.  Find  the  interest,  at  8%,  of  $13682.45,  borrowed  from 
a  minor  13  yr.  2  mo.  10  da.  old,  and  retained   till  he   is  of 
age  (21  years).  $8543.93 

24.  In  one  year  a  broker  loans  $876459.50  for  63  da.,  at 
1-|^£  a  mo.,  and  pays  6%  on  $106525.20  deposits:  what  is 
his  gain?  $21216.96 

25.  What  is  a  broker's  gain  in  1  yr.,  on  $100,  deposited 
at  6%,  and  loaned  11  times  for  33  da.,  at  the  rate  of  2% 
a  month?  $18.20 

26.  Find  the  simple  interest  of  £493  16s.   8d.  for  1  yr. 

8  mo.,  at  6^.  £49  7s.  8d. 
NOTE. — In  England  the  actual  number  of  days  is  counted. 

27.  Find  the  simple  interest  of  £24  18s.  9d.  for  10  mo., 
at  6%.  £1  4s.  Hid. 

28.  Of  £25  for  1  yr.  9  mo.,  at  5%.  £2  3s.  9d. 

29.  Of  £648   15s.  6d.   from  June   2  to  November  25,  at 
5%.  £15  12s.  lOd. 


250  'RAY1 8  HIGHER  ARITHMETIC. 


CASE    II. 

298.     Given  the  principal,  the  rate,  and  the  interest, 
to  find  the  time. 


PROBLEM.  —  John  Thomas  loaned  $480,  at  5%,  till  the  in- 
terest was  $150  ;  required  the  time. 

SOLUTION.  —  The    inter-  OPERATION. 

est  on  $480  for  1  yr.,  at  $  1  5  0  rf-  (  f  4  8  0  X  •  0  5  )  =  6  J  years. 
5^0,  is  $24.  If  the  prin- 

cipal produce  $24  interest  in  1  year,  it  will  require  as  many  years 
to  produce  $150  interest  as  $24  is  contained  times  in  $150,  which  is 
6g-  years,  or  6  yr.  3  mo.  From  the  preceding,  the  following  rule  is 
derived  : 

Kule.  —  Divide  the  given  interest  by  the  interest  of  the  prin- 
cipal for  one  year  ;  the  quotient  is  the  time. 

EEMARK.  —  If  the  principal   and   amount  are  given,  take   their 
difference  for  the  interest. 


EXAMPLES  FOR  PRACTICE. 

In  what  time  will : 

1.  $1200  amount  to  81800,  at  lO^g  ?  5  yr. 

2.  $415.50  to  $470.90,  at  10^?  1  yr.  4  mo. 

3.  $3703.92  to  $4122.15,  at  8^  ?  1  yr.  4  mo.  28  da. 

NOTE. — A  part  of   a  day,   not  being  recognized   in  interest,  is 
omitted  in  the  answer,  but  must  not  be  omitted  in  the  proof. 

4.  In  what  time  will  any  sum,  as  $100,  double  itself  by 
simple  interest,  at  4|,  6,  7£,  9,  10,  12,  20,  25,  30%  ? 

22|,  16?,  131,  Hi   10,  8J,  5,  4,  3£  yr. 


INTEREST.  251 

5.  In  what  time  will  any  sum  treble  itself  by  simple  in- 
terest, at  4,  10,  12%  ?  50,  20,  16|  yr. 

6.  How  long  must  I  keep  on  deposit  $1374.50,  at  10^, 
to  pay  a  debt  of  $1480.78?  9  mo.  8  da. 

7.  How  long  will  it  take  $3642.08  to  amount  to  $4007.54, 
at  12%?  10  mo.  1  da, 

8.  How  long  would  it  take  $175.12   to   produce   $6.43 
interest,  at  6%  ?  7  mo.  10  da. 

9.  How  long  would  it   take   $415.38  to  produce  $10.69 
interest  in  New  York,  at  l<fa  ?  134  days. 

CASE   III. 

299.     Given  the  principal,  interest,  and  time,  to  find 
the  rate. 

FORMULA.-  —=-  R 

-t  X  A  70  X  -L 

PROBLEM.  — At  what  rate  per  cent  will  $4800  gain  $840 
interest  in  2|-  years? 

SOLUTION. — The  OPERATION. 

interest   of   $4800,  Int.  of  $  4  8  0  0  at  1  ft  for  2  \-  yr.  =  $120; 

at  1^,  for  2Jyr.  is  /.     $840  -5-  $1  20  =  7  ;  or, 

$120 ;    hence,     the  1  ^,  X  7  =  7  ft,  Ansr 
rate    is    as    many 

times  \*f0  as  $120  is  contained  times  in  $840,  which  is.  7  times;  or, 
7^j.     Hence  the  rule. 

Rule. — Divide  the  given  interest  by  the  interest   of  the  prin- 
cipal for  the  given  time,  at  \%. 


EXAMPLES  FOR  PRACTICE. 

1.  At  what  rate  per  annum  will  any  sum  treble  itself,  at 
simple  interest,  in  5,  10,  15,  20,  25,  30  years,  respectively  ? 

40,  20,  131  10,  8,  6|#. 


252  RAY'S  HIGHER  ARITHMETIC. 

2.  At  what   rate   of  interest    per   annum   will   any  sum 
quadruple  itself,  at  simple  interest,  in  6,  12,  18,  24,  and  30 
years,  respectively?  50,  25,  16|,  12|,  10  ft. 

3.  What   is  the  rate  of  interest  when  $35000  yields  an 
income  of  8175  a  month?  6%. 

4.  What   is  the  rate  of  interest  when   $29200   produces 
$6. 40  a  day?  8%. 

5.  What   is  the    rate  of  interest   when   $12624.80  draws 
$315.62  interest  quarterly.  10%. 

6.  Find  the  rate  when   stock,  bought  at   40%    discount, 
yields  a  semi-annual  dividend  of  5%?         16f%  per  annum. 

7.  A   house  that  cost   $8250,  rents  for  $750  a  year;  the 
insurance  is  Ta0-^,  and  the  repairs  \%,   every  year:  what 
rate  of  interest  does  it  pay?  8% — . 

CASE    IV. 

300.     Given  the  interest,  rate,  and  time,  to  find  the 
principal. 

FORMULA.—  R^T  =  P. 

PROBLEM. — A  man  receives  $490.84  interest  annually  on 
a  mortgage,  at  7^5 :   what  is  the  amount  of  the  mortgage  ? 

SOLUTION. — Since   $.07    is  OPERATION. 

the  interest  of  $1  for  one  year,  $490.84^-$.  07  =  7012 

$490.84  is  the  interest  of  as  $1X7012  =  $7012,  Ans. 

many  dollars  as  $.07  is  con-  Or,           7  ^  —  $490.84 

tained  times  in  $490.84,  which  1^=$70.12 
is    7012  ;    therefore,  $701 2  is 
the  amount  of  the  mortgage. 

Or,  thus :  the  time  being  1  year,  $490.84  is  7^>  of  the  principal ; 

l<fo  of  it  is  \  of  $490.84,  which  is  $70.12,  and  hence  100^  of  it,  or  the 
whole  principal,  is  $7012. 

Rule. — Multiply  the  rate  by  the  time,  and  divide  the  interest 
by  the  product;   the  quotient  will  be  the  principal 


INTEREST. 


253 


EXAMPLES  FOR  PRACTICE. 

What  principal  will  produce: 

1.  $1500  a  year,  at  6%?  825000. 

2.  $1830  in  2  yr.  6  mo.,  at  5%  ?  $14640. 

3.  $45  a  mo.,  at  9^?  $6000. 

4.  $17  in  68  da.,  at  \%  a  month?  $750. 

5.  $656.25  in  9  mo.,  at  3£%  ?  $25000. 

6.  $86.15  in  9  mo.  11  da.,  at  10^?  $1103.70 

7.  $313.24  in  112  da.,  at  7%?  $14383.47 

8.  $146.05  in  7  mo.  14  da.,  at  6^?  $3912.05 

9.  $58.78  in  1  yr.  3  mo.  20  da.,  at  4%  ?  $1125.57 
10.  $79.12  in  5  mo.  25  da.,  at .1%  in  N.  Y.?  $2357.46 


CASE    V. 

301.     Given  the  amount,  rate,  and  time,  to  find  the 
principal. 

FORMULA.—  A  -T-  ( 1  +  K~><  T)  =  P. 

PROBLEM. — What  is  the   par  value  of  a  bond  which,  in 
8  yr.  8  mo.,  at  6^,  will  amount  to  $15200? 


OPERATION. 

5200-*-$!.  52  =  10000; 
Ans. 


SOLUTION. — There  are  as 
many  dollars  in  the  princi- 
pal as  $1.52  is  contained  Or,  100^  =  bond; 
times  in  $15200,  which  is 
10000;  therefore,  $10000  is 
the  par  value  of  the  bond. 
Or,  the  amount  is  -J-^J  of 

the  principal ;  hence,  ^f§  of  the  principal  is  $15200 ;  and 
principal  is  $100;  and  -j^g§  of  the  principal  is  $10000. 


100  $,=r$  10000,  Am. 


of  the 


Rule. — Divide    the    amount    by  the    amount    of  $1  for  the 
given  time  and  rate;   the  quotient  will  be  the  principal. 


254  RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  What   principal  in  2   yr.    3   mo.  12  da.,  at  6^,  will 
amount  to  $1367.84?  $1203.03 

2.  What    principal   in   10  mo.    26    da.,   will  amount  to 
$2718.96,  at  10%  interest?  $2493.19 

3.  What  principal,  at  4^%,  will  amount  to  $4613.36  in 
3  yr.  1  mo.  7  da.?  $4048.14 

4.  What  principal,  at  7%,  will  amount  to  $562.07  in  79 
da.  (365  da.  to  a  year)  ?  $553.68 


PKOMISSOKY  NOTES. 
DEFINITIONS. 

302.     1.  A  Promissory  Note   is  a  written  promise  by 
one  party  to  pay  a  specified  sum  to  another. 

2.  The  Face  of  a  note  is  the  sum  promised  to  be  paid. 

3.  The  Maker  is  the  party  who  binds  himself  to  pay  the 
note  by  signing  his  name  to  it. 

4.  The  Payee    is  the  party   to    whom    the  payment  is 
promised. 

NOTE. — The  maker  and  the  payee  are  called  the  original  parties  to 
a  note. 

5.  The  Holder  of  a  note  is  the  person  who  owns  it:  he 
may  or  may  not  be  the  payee. 

6.  The  Indorser  of  a  note   is  the   person  who  writes  his 
name  across  the  back  of  it. 

7.  A  Time  Note  is  one  in  which  a  specified  time  is  set 
for  payment. 

REMARK. — There  are  various  forms  used  in  time  notes,  the  prin- 
cipal ones  being  as  follows: 


PR OMISSOE  Y  NOTES.  255 

1.  ORDINARY  FORM. 

$450.50  BOSTON,  MASS.,  June  30,  1880. 

Three  montlis  after  date,  I  promise  to  pay  Thomas  Ford, 
or  order,  Four  Hundred  and  Fifty,  and  -ffa,  Dollars,  for  value 

received,  with  interest. 

EDWARD  E.  MORRIS. 

2.  JOINT  NOTE. 

$350.  DENVER,  COL.,  Jan.  2,  1881. 

Twelve  months  after  date,  we,  or  either  of  us,  promise  to 
pay  Frank  R.  Harris,  or  bearer,  Three  Hundred  and  Fifty 
Dollars,  for 'value  received,  with  interest  from  date. 

JAMES  WEST. 
DANIEL  TATE. 

REMARKS. — 1.  This  note  is  called  a  "joint  note,"  because  James 
West  and  Daniel  Tate  are  jointly  liable  for  its  payment. 

2.  If  the  note  read,  "  We  jointly  and  severally  promise  to  pay," 
etc.,  it  would  then  be  called  a  joint  and  several  note,  and  the  makers 
would  be  both  jointly  and  singly  liable  for  its  payment. 

3.  PRINCIPAL  AND  SURETY  NOTE. 

$125.00  ST.  Louis,  Mo.,  Nov.  20,  1880. 

Ninety  days  after  date,  I  promise  to  pay  James  Miller,  or 
order,  One  Hundred  and  Twenty-jive  Dollars,  for  value  received, 
negotiable  and  payable  without  defalcation  or  discount. 

Surety,  JAMES  MILLER.  H.  A.  WHITE. 

REMARK. — The  maker  of  this  note,  H.  A.  White,  is  the  principal, 
and  the  note  is  made  payable  to  the  order  of  the  surety,  James  Miller, 
who  becomes  security  for  its  payment,  indorsing  it  on  the  back  to 
the  order  of  Mr.  White's  creditor. 


256  JRAY'S  HIGHER  ARITHMETIC. 

8.  A  Demand  Note  is  one  in  which  no  time  is  specified, 
and  the  payment  must  be  made  whenever  demanded  by  the 
holder.  The  following  is  an  example : 


DEMAND   NOTE. 
$1100.00  CINCINNATI,  O.,  Mar.  18,  1881. 

For  value  received,  I  promise  to  pay  David  Swinton,  or 
order,  on  demand,  Eleven  Hundred  Dollars,  with  interest. 

HENRY  KUDOLPH. 

9.  A  promissory  note  is  negotiable — that  is,  transferable 
to   another   party   by   indorsement, — when    the    words    "or 
order,"  or  "or  bearer,"  follow  the  name  of  the  payee,  as  in 
the  above  examples ;  otherwise,  the  note  is  not  negotiable. 

10.  If  the  note  is  payable  to  "bearer,"  no  indorsement  is 
required  on  transferring  it  to  another  party,  and  the  maker 
only  is  responsible  for  its  payment. 

11.  If  the  note  is  payable  to  "order,"  the  payee  and  each 
holder  in  his  turn  must,  on  transferring  it,  indorse  the  note 
by  writing  his  name  on  its  back,  thus  becoming  liable  for  its 
payment,  in  case  the  maker  fails  to  meet  it  when  due. 

REMARKS. — 1.  An  indorser  may  free  himself  from  responsibility 
for  payment  if  he  accompany  his  signature  with  the  words,  "with- 
out recourse ;"  in  which  case  his  indorsement  simply  signifies  the 
transfer  of  ownership. 

2.  If  the  indorser  simply  writes  his  name  on  the  back  of  the  note, 
it  is  called  a  "  blank  "  indorsement ;  but  if  lie  precedes  his  signature 
by  the  words,  "  Pay  to  the  order  of  John  Smith,"  for  example,  it  is 
called  a  "  special "  indorsement,  and  John  Smith  must  indorse  the 
note  before  it  can  legally  pass  to  a  third  holder. 

12.  It    is  essential    to    a  valid    promissory  note,   that  it 
contain  the  words   "value  received,"  and  that  the  sum  of 
money  to  be  paid  should  be  written  in  words. 


PR OMISSOR  Y  NOTES.  257 

13.  If  a  note  contains  the  words  "with  interest,"  it  draws 
interest  from  date,    and,  if  no  rate  is  mentioned,   the  legal 
rate  prevails. 

REMARKS. — 1.  The  face  of  such  a  note  is  the  sum  mentioned  with 
its  interest  from  date  to  the  day  of  payment. 

2.  If  a  note  does  not  contain  the  words  "  with  interest,"  and  is 
not  paid  when  due,  it  draws  interest  from  the  day  of  maturity,  at 
the  legal  rate,  till  paid. 

14.  A  note  matures  on  the  day  that  it  is  legally  due. 
However,  in  many  of  the  states,  three  days,  called  Days  of 
Grace,  are  allowed  for  the  payment  of  a  note   after  it  is 
nominally  due. 

REMARKS. — 1.  The  day  before  "grace"  begins,  the  note  is  nomi- 
nally due  ;  it  is  legally  due  on  the  last  day  of  grace. 

2.  The  days  of  grace  are  rejected  by  writing  "  without  grace  "  in 
the  note. 

3.  Notes  falling  due  on  Sunday  or  on  a  legal  holiday,  are  due  the 
day  before  or  the  day  after,  according  to  the  special  statute  of  each 
state  or  territory. 

4.  If  a  note  is  payable  "on  demand,"  it  is  legally  due  when  pre- 
sented. 

15.  When  a  note  in  bank  is  not  paid  at  maturity,  it  goes 
to  protest — that  is,  a  written  notice  of  this  fact,  made  out 
in  legal  form  by  a  notary  public,  is  served  on  the  indorsers, 
or  security. 

REMARK. — Some  of  the  states  require  that  certain  words  shall  be 
inserted  in  every  negotiable  note  in  addition  to  the  usual  forms. 
For  instance,  in  Pennsylvania,  the  words  "  without  defalcation  "  are 
required ;  in  New  Jersey,  "  without  defalcation  or  discount ;"  and 
in  Missouri,  the  statute  requires  the  insertion  of  "  negotiable  and 
payable  without  defalcation  or  discount."  In  Indiana,  in  order  to 
evade  certain  provisions  of  the  law,  the  following  words  are  usually 
inserted,  "without  any  relief  from  valuation  or  appraisement  laws." 
This  constitutes  what  is  known  as  the  "iron-clad"  note. 

303.     If  a  note  be  payable  a  certain  time  "  after  date," 

proceed  thus  to  find  the  day  it  is  legally  due: 
H.  A.  22. 


258 


RAY'S  HIGHER  ARITHMETIC. 


Rule. — Add  to  the  date  of  the  note,  the  number  of  years  and 
months  to  elapse  before  payment ;  if  this  gives  the  day  of  a  month 
higher  than  that  month  contains,  take  the  last  day  in  that  month; 
then,  count  the  number  of  days  mentioned  in  the  note  and  3  more : 
this  will  give  the  day  the  note  is  legally  due ;  but  if  it  is  a  Sunday 
or  a  national  holiday,  it  must  be  paid  the  day  previous. 

REMARKS.—!.  When  counting  the  days,  do  not  reckon  the  one  from 
which  the  counting  begins.  (For  exceptions,  see  Art.  313, 16,  Rem.2.) 

2.  The  months  mentioned  in  a  note  are  calendar  months.     Hence, 
a  3  mo.  note  would  run  longer  at   one   time  than  at  another ;  one 
dated  Jan.  1st,  will  run  93  days ;  one  dated  Oct.  1st,  will  run   95 
days :  to  avoid  this  irregularity,  the  lime  of  short  notes  is  generally 
given  in  days  instead  of  months ;  as,  30,  60,  90  days,  etc. 

3.  When  the  time  is  given  in  days,  it  is  convenient  to  use  the 
following  table  in  determining  the  day  of  maturity  of  a  note  : 


TABLE 

Showing  the  Number  of  Days  from  any  Day  of: 


*H 

P 

•3 

Or* 

K 
P 
N 

> 
*& 
*t 

g 
P 
*4 

«H 

5 
<t> 

«H 

c 

<ST 

> 

CTQ 

% 
HI 

£ 

e» 

4 

o 

<! 

$ 

p 

To  the 
same  day 
of  next 

365 

334 

306 

275 

245 

214 

184 

153 

122 

92 

61 

31 

Jan. 

31 

365 

337 

306 

276 

245 

215 

184 

153 

123 

92 

62 

Feb. 

59 

28 

365 

334 

304 

273 

243 

212 

181 

151 

120 

90 

Mar. 

90 

59 

31 

365 

335 

304 

274 

243 

212 

182 

151 

121 

April. 

120 

89 

61 

30 

365 

334 

304 

273 

242 

212 

181 
212 

151 

Mav. 

151 

120 

92 

61 

31 

365 

335 

304 

273 

243 

182 

June. 

181 

150 

122 

91 

61 

30 

365 

334 

303 

273 

242 

212 

July. 

212 

181 

153 

122 

92 

61 

31 

365 

334 

304 

273 

243 

Aug. 

243 

212 

184 

153 

123 

92 

62 

31 

365 

30 

335 

304 

274 

Sept. 

273 

242 
"273 

214 

183 

153 

122 

92 

61 

365 
31 

334 

304 

Oct. 

304 

245 

214 

184 

153 

123 

92 

61 

365 

335 

Nov. 

334 

303 

275 

244 

214 

183 

153 

122 

91 

61 

30 

365 

Dec. 

REMARK. — In  leap  years,  if  the  last  day  of  February  be  included 
in  the  time,  1  day  must  be  added  to  the  number  obtained  from  the 
table. 


ANNUAL  INTEREST.  259 


EXAMPLES  FOR  PRACTICE. 

Find  the  maturity  and  amount  of  each  of  the  following 
notes : 

$560.60  CHATTANOOGA,  TENN.,  June  3,  1872. 

1.  For  value  received,  sixty  days  after  date,  I  promise  to 
pay  Madison   Wilcox  five  hundred   and  sixty  y6^  dollars, 
with  interest  at  7^.  JAMES  DAILY. 

$567.47— 

$430.00  TOPEKA,  KAN.,  Jan.  30,  1879. 

2.  Six  months  after  date,  I  promise  to  pay  Christine  Ladd 
four  hundred   and  thirty  dollars,   for  value  received,  with 
interest  at  12%  per  annum.  AMOS  LYLE. 

$456.23 

$4650.80  ST.  Louis,  Mo.,  August  10,  1875. 

3.  For  value  received,  three  months  after  date,  I  promise 
to  pay  Oliver  Davis,  or  order,  four   thousand  six  hundred 
and  fifty  y8^-  dollars,  with  interest  at  the  rate  of  10%  per 
annum,  negotiable  and   payable  without  defalcation  or  dis- 
count. MILTON  MOORE. 

Surety,  OLIVER  DAVIS.  $4770.95 — 


ANNUAL   INTEKEST. 

304.  Annual  Interest  is  interest  on  the  principal,  and 
on  each  annual  interest  after  it  is  due. 

Annual  interest  is  legal  in  some  of  the  states. 

EXPLANATION.— If  Charles  Heydon  gives  a  note  and  agrees  to  pay 
the  interest  annually,  but  fails  to  do  so,  and  lets  the  note  run  for 
three  years  before  settlement,  the  first  year's  interest  would  draw 
interest  for  two  years,  and  the  second  year's  interest  would  draw 
interest  for  one  year.  The  last  year's  interest  would  be  paid  at  the 
time  of  settlement,  or  as  it  fell  due. 


260  RAY'S  HIGHER  ARITHMETIC. 

PROBLEM. — Find  the  amount  of  $10500  for  4  yr.  6  mo., 
interest  6^,  payable  annually. 

OPERATION. 

Int.  of  $  1  0  5  0  0  for  4  J  yr.,  at  6fi  =  $  2  8  3  5; 
"      "    $630          "8      "      ""=$302.40 


Total  annual  interest,  =$3137.40 

Amount  =$10500 +  $3  137.40  =  $!  36  3  7. 40 

SOLUTION.— The  interest  of  $10500  for  1  yr.,  at  6^,  is  $630,  and 
for  4J  yr.  is  $2835.  The  first  annual  interest  draws  interest  3J  yr., 
the  second  2J,  the  third  1J,  and  the  fourth  \  yr.  This  is  the  same 
as  one  annual  interest  for  (3|  +  2J  +  1|  +  -J-  —  8)  eight  years.  But 
the  interest  of  $630,  for  8  yr.,  at  6^>,  is  $302.40;  therefore,  the 
amount  ==  $10500  +  $2835  +  $302.40  =  $13637.40 

Rule. — 1.  Find  the  interest  on  the  principal  for  the  entire 
time,  and  on  each  year's  interest  till  the  time  of  settlement. 
2.   The  sum  of  these  interests  is  the  interest  required. 

NOTE.— In  Ohio  and  several  other  states,  interest  on  unpaid  annual 
interest  is  calculated  at  the  legal  rate  only. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  amount  of  a  note   for  $1500,  interest    6%, 
payable   annually,  given   Sept.   3,   1870,   and   not   paid   till 
March  1,  1874.  $1838.71 

2.  A  gentleman  holds  six  $1000  railroad  bonds,  due  in 
3  years,    interest  6^ ,    payable    semi-annually :    no    interest 
having   been    paid,  what  amount    is   owing   him   when   the 
bonds  mature?  $7161. 

3.  $2500.00  ST.  PAUL,  MINN.,  Jan.  11,  1869. 
For  value  received,   I  promise   to  pay  Morgan  Stuart, 

or  order,  twenty-five   hundred  dollars,  with  interest  at  7%, 
payable  annually.  LEONARD  DOUGLAS. 

What  was  due  on  this  note  March  17,  1873,  if  the  interest 
was  paid  the  first  two  years?  $2898.825 


PARTIAL  PAYMENTS.  261 


II.   PARTIAL   PAYMENTS. 

305.  A  Partial  Payment  is  a  payment  in  part  of  a 
note  or  other  obligation. 

Whenever  a  partial  payment  is  made,  the  holder  should 
write  the  date  and  amount  of  the  payment  on  the  back  of 
the  note.  This  is  called  an  Indorsement. 

308.  The  following  decision  by  Chancellor  Kent,  "  John- 
son's Chancery  Rep.,  Vol.  I,  p.  17,"  has  been  adopted  by  the 
Supreme  Court  of  the  United  States,  and,  with  few  excep- 
tions, by  the  several  states  of  the  Union,  and  is  known  as 
the  "  United  States  Rule:" 

U.  S.  Rule. — 1.  "The  rule  for  casting  interest  when  par- 
tial payments  have  been  made,  is  to  apply  the  payment,  in  the 
first  place,  to  the  discharge  of  the  interest  due. 

2.  "If  the  payment    exceeds    the    interest,    the   surplus   goes 
towards  discharging  the  principal,  and   the   subsequent   interest 
is  to  be  computed  on  the  balance  of  principal  remaining  due. 

3.  "If  the  payment   be  less  than  the  interest,  the  surplus  of 
interest   must  not   be  taken  to  augment   the  principal;    but  in- 
terest continues  on  the  former  principal   until  the  period  when 
the  payments,  taken    together,  exceed  the  interest  due,  and  then 
the  surplus  is  to  be  applied    towards  discharging  the  principal, 
and  interest  is  to  be  computed  on  the  balance  as  aforesaid. 

307.     This  rule  is  based  upon  the  following  principles: 

PRINCIPLES. — 1.  Payments  must  be  applied  first  to  the  dis- 
charge of  interest  due,  and  the  balance,  if  any,  toward  paying 
the  principal. 

2.  Interest  must,  in  no  case,  become  part  of  the  principal. 

3.  Interest  or  payment  must  not  draw  interest. 

REMARKS. — 1.  It  is  worthy  of  remark,  that  the  whole  aim  and 
tenor  of  legislative  enactments  and  judicial  decisions  on  questions  of 


262  RAY'S  HIGHER  ARITHMETIC. 

interest,  have  been  to  favor  the  debtor,  by  disallowing  ^compound 
interest,  and  yet  this  very  rule  fails  to  secure  the  end  in  view,  and 
really  maintains  and  enforces  the  principle  of  compound  interest  in 
a  most  objectionable  shape ;  for  it  makes  interest  due  (not  every  year  as 
compound  interest  ordinarily  does),  but  as  often  as  a  payment  is  made;  by 
which  it  happens  that  the  closer  the  payments  are  together,  the  greater  the 
loss  of  the  debtor,  who  thus  suffers  a  penalty  for  his  very  promptness. 

To  illustrate,  suppose  the  note  to  be  for  $2000,  drawing  interest 
at  6^,  and  the  debtor  pays  every  month  $10,  which  just  meets  the 
interest  then  due;  at  the  end  of  the  year  he  would  still  owe  $2000. 
But  if  he  had  invested  the  $10  each  month,  at  6^,  he  would  have 
had,  at  the  end  of  the  year,  $123.30  available  for  payment,  while  the 
debt  would  have  increased  only  $120,  being  a  difference  of  $3.30  in 
his  favor,  and  leaving  his  debt  $1996.70,  instead  of  $2000. 

2.  To  find  the  difference  of  time  between  two  dates  on  the  note, 
reckon  by  years  and  months  as  far  as  possible,  and  then  count  the 
days. 

PROBLEM. 


CINCINNATI,  April  29, 1880. 
Ninety   days    after    date,   I    promise   to    pay    Stephen 
Ludlow,  or  order,  eight  hundred  and  fifty  dollars,  with  in- 
terest; value  received.  CHARLES  K.  TAYLOR. 

Indorsements.— Oct.  13, 1880,  $40 ;  June  9, 1881,  $32  ;  Aug.  21, 1881, 
$125 ;  Dec.  1,  1881,  $10 ;  March  16,  1882,  $80. 

What  was  due  Nov.  11,  1882? 

SOLUTION. — Interest  on  principal  ($850)  from  April  29 

to  Oct.  13,  1880,  5  mo.  14  da.,  at  6$  per  annum,  $23.233 

850. 

Whole  sum  due  Oct.  13,  1880,        .  873.233 

Payment  to  be  deducted,  40. 

Balance  due  Oct.  13,  1880,               .  833.233 
Interest  on   balance    ($833.233)  from  Oct.  13,  1880,  to 

June  9,  1881,  being  7  mo.  27  da.,      0  32.913 

Payment  not  enough  to  meet  the  interest,       .        .         .  32. 

Surplus  interest  not  paid  June  9,  1881,  ....  ^913 
Interest  on  former  principal  ($833.233)  from  June  9, 

1881,  to  Aug.  21,  1881,  being  2  mo.  12  da,,       .        .  9.999 

Whole  interest  due  Aug.  21,  1881, 10.912 


PARTIAL  PAYMENTS.  263 

(Brought  forward)  Int.  due  Aug.  21,  1881,    .                 .  10.912 

833.233 

Whole  sum  due  Aug.  21,  1881, 844.145 

Payment  to  be  deducted, 125. 

Balance  due  Aug.  21,  1881, 719.145 

Interest  on  the  above  balance  ($719.145)  from  Aug.  21, 

1881,  to  Dec.  1,  1881,  being  3  mo.  10  da.,        .        .  11.986 

Payment  not  enough  to  meet  the  interest,       .         .        .  10. 

Surplus  interest  not  paid  Dec.  1,  1881,     ,  1.986 
Interest  on  former  principal    ($719.145)   from   Dec.  1, 

1881,  to  March  16,  1882,  being  3  mo.  15  da.,    .         .  12.585 

Whole  interest  due  March  16,  1882,        ....  14.571 

719.145 

Whole  sum  due  March  16,  1882,      .        .        .        .        .  733.716 

Payment  to  be  deducted,  ...         t         ...  80. 

Balance  due  March  16,  1882, 653.716 

Interest  on  balance  ($653.716)  from  March  16,  1882,  to 

Nov.  11,  1882,  being  7  mo.  26  da.,     ....  25.713 
Balance  due  on  settlement,  Nov.  11,  1882,       .        .        .  $679.43 


EXAMPLES  FOR  PRACTICE. 

1.  $304T7^-.  CHICAGO,  March  10,  1882. 

For  value  received,  six  months  after  date,  I  promise  to 
pay  G.  Riley,  or  order,  three  hundred  and  four  T7^-  dollars. 

No  payments.  H.   McMAKIN. 

What  was  due  Nov.  3.  1883?  $325.63 

2.  S429y\V  INDIANAPOLIS,  April  13,  1873. 

On  demand,  I  promise  to  pay  W.  Morgan,  or  order, 
four  hundred  and  twenty-nine  ffa  dollars,  value  received, 
with  interest.  R  WlLSON. 

Indorsed:    Oct.  2,1873,  $10;    Dec.  8,  1873,  $60;   July  17,  1874, 

$200. 

What  was  due  Jan.  1,  1875?  $195.06 


264  RAY'S  HIGHER  ARITHMETIC. 

3.  $1750.  NEW  YORK,  Nov.  22,  1872. 
For  value  received,  two  years  after  date,  I  promise  to 

pay  to  the  order  of  Spencer  &  Ward,  seventeen  hundred 
and  fifty  dollars,  with  interest  at  7  per  cent.' 

JACOB  WINSTON. 

Indorsed:  Nov.  25,  1874,  $500;  July  18,  1875,  $50;  Sept.  1,  1875, 
$600 ;  Dec.  28,  1875,  $75. 

What  was  due  Feb.  10,  1876?  $879.71 

4.  A  note  of  $312  given  April  1,  1872,  8%  from  date, 
was  settled  July  1,  1874,  the  exact  sum  due  being  $304.98. 

Indorsed:    April   1,  1873,  $30.96;   Oct.  1,  1873,$ ;    April  1, 

1874,  $20.40 

Restore  the  lost  figures  of  the  second  payment.  $11.08 

5.  There  have  been  two  equal  annual  payments  on  a  6^ 
note  for  $175,  given  two  years  ago  this  day.     The  balance  is 
£154.40:  what  was  each  payment?  $20.50 

6.  A    merchant    gives    his    note,    10%    from    date,    fov 
$2442.04:    what  sum   paid  annually  will    have    discharged 
the   whole   at  the  end  of  5   years?  $644.204 

308.  In  Connecticut  and  Vermont  the  laws  provide  the 
following  special  rules  relative  to  partial  payments: 

Connecticut  Rule. — 1.  Compute  the  interest  to  the  time  of 
the  first  payment,  if  that  be  one  year  or  more  from  the  time 
that  interest  commenced;  add  it  to  the  principal,  and  deduct 
the  payment  from  the  sum  total. 

2.  If  there  be  after  payments  made,  compute  the  interest  on 
the  balance  due  to  the  next  payment,  and  then  deduct  the  pay- 
ment   as    above;    and    in    like    manner  from  one   payment  to 
another,  till  all  the  payments  are  absorbed:  PROVIDED,  the  time 
between  one  payment  and  another  be  one  year  or  more. 

3.  BUT    if  any  payment  be  made  before   one  year's  interest 
hath  accrued,  then  compute  the  interest,  on   the  sum   then  due 
on   the   obligation,  for  one  year,  add   it  to  the  principal,  and 
compute  the   interest  on  the  sum  paid,  from  the  time  it  was 


PARTIAL  PAYMENTS.  265 

paid,  up  to  the  end  of  the  year;  add  it  to  the  sum  paid,  and 
deduct  that  sum  from  the  principal  and  interest  added  as  above. 
(See  Note.) 

4.  If  any  payment  be  made  of  a  less  sum  than  the  interest 
arisen  at  the  time  of  such  payment,  no  interest  is  to  be  com- 
puted, but  only  on  the  principal  sum,  for  any  period. 

NOTE. — If  a  year  does  not  extend  beyond  the  time  of  payment;  but 
if  it  does,  then  find  the  amount  of  the  principal  remaining  unpaid  up  to 
the  time  of  settlement,  likewise  the  amount  of  the  payment  or  payments 
from  the  time  they  were  paid  to  the  time  of  settlement,  and  deduct  the  sum 
of  these  several  amounts  from  the  amount  of  the  principal. 

What  is  due  on  the  2d  and  3d  of  the  preceding  notes,  by 
the  Connecticut  rule?  2d,  $194.88  ;  3d,  $877.95 

Vermont  Rule. — 1.  On  all  notes,  etc.,  payable  WITH  IN- 
TEREST, partial  payments  are  applied,  first,  to  liquidate  the 
interest  that  has  accrued  at  the  time  of  such  payments;  and, 
secondly,  to  the  extinguishment  of  the  principal. 

2.  When  notes,  etc.,  are  drawn  with   INTEREST    PAYABLE 
ANNUALLY,  the  annual  interests  that  remain  unpaid  are  subject 
to  simple  interest  from  the  time  they  become  due  to  the  time 
of  final  settlement. 

3.  PARTIAL  PAYMENTS,   made  after  annual  interest  begins 
to  accrue,  ALSO  DRAW  SIMPLE  INTEREST  from  the  time  such 
payments  are  made  to  the  end  of  the  year;  and  their  amounts 
are  then  applied,  first,  to  liquidate  the  simple  interest  that  has 
accrued  from   the   unpaid   annual  interests;   secondly,  to  liqui- 
date  the   annual  interests   that  have  become  due;   and,  thirdly, 
to  the  extinguishment  of  the  principal. 

$1480.  WOODSTOCK,  VT.,  April  12,  1879. 

For  value  received,  I  promise  to  pay  John  Jay,  or  order, 
fourteen  hundred  and  eighty  dollars,  with  interest  annually. 

JAMES  BROWN. 

Indorsed  :  July  25, 1879,  $40 ;  May  20, 1880,  $50 ;  June  3, 1881,  $350. 
What  was  due  April  12,  1882?  $1291.95 

H.  A.  23. 


266  RAY'S  HIGHER  ARITHMETIC. 

309.  Business    men  generally  settle  notes  and  accounts 
payable  within  a  year,  by  the  Mercantile  Rule. 

Mercantile  Rule. — 1.  Find  the  amount  of  the  principal 
from  the  date  of  the  note  to  the  date  of  settlement. 

2.  Find   the  amount  of  each  payment  from   its  date  to  the 
date  of  settlement. 

3.  From  the  amount  of  the  principal  subtract  the  sum  of  the 
amounts  of  the  payments. 

NOTE. — In  applying  this  rule,  the  time  should  be  found  for  the 
exact  number  of  days. 

EXAMPLES  FOR  PRACTICE. 

1.  $950.00  NEW  YORK,  Jan.  25,  1876. 
For  value  received,  nine  months  after  date,  I  promise 

to  pay  Peter  Finley  nine  hundred  and  fifty  dollars,  with  1% 
interest.  THOMAS  HUNTER. 

The  following  payments  were  made  on  this  note:  March 
2,  1876,  $225;  May  5,  1876,  $174.19;  June  29,  1876, 
$187.50;  Aug.  1,  1876,  $79.15 

Required — the  balance  at  settlement.  $312.57 

2.  A  note  for  $600  was  given  June  12,  1865,  6%  interest, 
and  the  following  indorsements  were  made :  Aug.  12,  1865, 
$100;  Nov.  12,  1865,  $250;  Jan.  12,  1866,  $120:  what  was 
due  Feb.  12,  1866,  counting  by  months  instead  of  days  ? 

$146.65 

III.    TKUE   DISCOUNT. 
DEFINITIONS. 

310.  1.  Discount  on   a  debt  payable  by  agreement  at 
some  future  time,  is  a  deduction  made  for  "  cash,"  or  present 
payment. 

It  arises  from  the  consideration  of  the  present  worth  of  the 
debt. 


TRUE  DISCOUNT.  267 

2.  Present    Worth    is    that  sum   of  money  which,  at  a 
given  rate  of  interest,  will  amount  to  the  same  as  the  debt 
at  its  maturity. 

3.  True   Discount,  then,    is    the  difference   between  the 
present  worth  and  the  whole  debt. 

REMARKS. — 1.  That  is,  it  is  the  simple  interest  on  the  present 
worth,  from  the  day  of  discount  until  the  day  of  maturity. 

2.  Discount  on  a  debt  must  be  carefully  distinguished  from  Com- 
mercial Discount,  which  is  simply  a  deduction  from  the  regulai 
price  or  value  of  an  article ;  the  latter  is  usually  expressed  as  such 
a  "  per  cent  off." 

311.  The  different  cases  of  true  discount  may  be  solved 
like  those  of  simple  interest:  the  present  worth  correspond- 
ing to  the  Principal;   the  discount,  to  the  Interest;   and  the 
face  of  the  debt,    to    the  Amount.     The    following  case    is 
the  only  one   much  used : 

312.  Given   the   face,  time,   and    rate,   to    find   the 
present  worth  and  true  discount. 

PROBLEM. — Find  the  present  worth  and  discount  of 
$5101.75,  due  in  1  yr.  9  mo.  19  da.,  at  6  ft. 

OPERATION. 

Amount  of  $  1  for  1  yr.  9  mo.  1  9  da.,  at  6  <fc  —  $  1 .  l  Q  8  £, 
and  $ 5 1 0 1 . 7 5 -s- $  1 . 1 0 8  J  =  46 0 3 , 7 T 5 j 

$1X4603.  77  5  =  $4603.  77  5,  present  worth ; 
$5101.75—  $4603. 77  5  =$497. 97  5,  true  discount. 

SOLUTION. — The  amount  of  $1  for  1  yr.  9  mo.  19  da.,  at  6^,  is 
$1.108J,  and  $5101.75  is  the  amount  of  as  many  dollars  as  $1.108£  is 
contained  times  in  $5101.75,  which  is  4603.775  times ;  therefore,  the 
present  worth  is  $4603.775,  and  the  true  discount  is  $497.975 

Rule. — 1.  Divide  the  debt  by  the  amount  of  $1  for  the  given 
time  and  rate ;  the  quotient  is  the  present  worth. 

2.  The  difference  between  the  debt  and  the  present  worth  is 
the  true  discount.  ^ 

OF  THE 


268  PA  Y '  S  HIGHER   ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  true  discount  on  a  debt  for  $5034.15  due  in 
3  yr.  5  mo,  20  da.,  without  grace,  at  1%.  .    $984.33 

2.  What  is  the  present  worth  of  a  note  for  $2500,  bearing 
6%  interest,  and  payable  in  2  yr.  6  mo.  15  da.,  discounted 
at  8%.,  $2394.10 

IV.    BANK   DISCOUNT. 

DEFINITIONS. 

313.  1.  A  Bank  is  an  institution  authorized  by  law  to 
deal  in  money. 

2.  The  three  chief  provinces  of  banks  are :  the  receiv- 
ing of  deposits,  the  loaning  of  money,  and  the  issuing  of 
bank-bills.  Any  or  all  of  these  provinces  may  be  exercised 
by  the  same  bank. 

REMARKS. — 1.  A  Bank  of  Deposit  is  one  which  takes  charge  of 
money,  stocks,  etc.,  belonging  to  its  customers. 

Money  so  intrusted  is  called  a  deposit,  and  the  customers  are 
known  as  depositors. 

2.  A  Sank  of  Discount  is  one  which  loans  money  by  discounting 
notes,  drafts,  etc. 

3.  A  Bank  of  Issue  is  one  which  issues  notes,  called  "  bank-bills," 
that  circulate  as  money. 

.     3.  The  banks  of  the.  United  States  may  be  divided   into 
two  general  classes, — National  Banks  and  Private  Banks. 

REMARKS. — 1.  National  Banks  are  organized  under  special  legis- 
lation of  Congress.  They  must  conform  to  certain  restrictions  as  to 
the  number  of  stockholders,  amount  of  capital  stock,  reserve,  circu- 
lation, etc.  In  return,  they  have  privileges  which  give  them  certain 
advantages  over  Private  Banks  :  they  are  banks  of  issue,  of  discount, 
and  of  deposit. 

2.  Private  Banks  are  unable  to  issue  their  own  bank-notes  to 
advantage,  owing  to  the  heavy  tax  on  their  circulation  ;  they  are, 
therefore,  confined  to  the  provinces  of  deposit  and  discount. 


BANK  DISCOUNT.  269 

4.  A  Check  is  a  written  order  on  a  bank  by  a  depositor 
for  money.     The  following  is  a  usual  form: 


No.   1032.  CINCINNATI,  Nov.  27,  1881. 

BANK   OF    CINCINNATI, 

PAY  TO Rufus  B.  Shaffer, OR  ORDER 

Four  Hundred  and  Twenty-five T£-g-  DOLLARS. 

$425^  George  Potter  Hottister. 


REMARKS.—!.  Before  this  check  is  paid,  it  must  be  indorsed  by 
Rufus  B.  Shaffer.  He  may  either  draw  the  money  from  the  bank 
itself,  or  he  may  transfer  the  check  to  another  party,  who  must  also 
indorse  it,  as  in  the  case  of  a  promissory  note.  (See  Art.  302,  11.) 

2.  If  the  words  or  Bearer  are  used  in  place  of  or  Order,  no  indorse- 
ment is  necessary,  and  any  one  holding  the  check  may  draw  the 
money  for  it. 

5.  A  Draft  is  a  written  order  of  one  person  or  company 
upon  another  for  the  payment  of  money.  The  following  is 
a  customary  form: 


$1453^.  NEW  ORLEANS,  LA.,  July  25,  1880. 

. L At  sigJd PAY  TO  THE 

ORDER  OF First  National  Bank 


Fourteen  Hundred  Fifty-three  and y^-g-  DOLLARS 

VALUE  RECEIVED  AND  CHARGE  TO  ACCOUNT  OF 

To    John  R.    Williams  &  Co.,    j          Roberi  Jar)W§f 
No.  2136.  Memphis,   Tenn.  \ 


270  RA  Y>  S  HIGHER  ARITHMETIC. 

6.  Drafts  may  be  divided  into  two   classes,  Sight  Drafts 
and  Time  Drafts. 

7.  A  Sight  Draft  is  one  payable  "at  sight."     (See  the 
form  above.) 

8.  A  Time    Draft   is   payable    a    specified    time   "after 
sight,"  or  "after  date." 

9.  The  signer  of  the  draft  is  the  maker  or  drawer. 

10.  The  one  to  whom  the  draft  is  addressed,  and  who  is 
requested  to  pay  it,  is  the  drawee. 

11.  The  one  to  whom  the  money  is  ordered  to  be  paid,  is 
the  payee. 

12.  The  one  who  has  possession  of  the  draft,  is  called 
the   owner  or  holder;   when  he   sells  it,   and   becomes  an 
indorser,  he  is  liable  for  its  payment. 

13.  The  Indorsement  of  a  draft  is  the  writing  upon  the 
back  of  it,  by  which  the  payee  transfers  the   payment   to 
another. 

REMARKS. — 1.  A  special  indorsement  is  an  order  to  pay  the  draft  to 
a  particular  person  named,  who  is  called  the  indorsee,  as  "Pay. to 
F.  H.  Lee. — W.  Harris,"  and  no  one  but  the  indorsee  can  collect  the 
bill.  Drafts  are  usually  collected  through  banks. 

2.  When  the  indorsement  is  in  blank,  the  payee  merely  writes  his 
name  on  the  back,  and   any  one   who  has  lawful  possession  of  the 
draft  can  collect  it. 

3.  If  the  drawee  promises  to  pay  a  draft  at  maturity,  he  writes 
across  the  face  the  word  "Accepted,"  with  the  date,  and  signs  his 
name,  thus:  "Accepted,  July  11,  1881. — H.  Morton."      The  acceptor 
is  first  responsible  for  payment,  and  the  draft  is  called  an  acceptance. 

4.  A  draft,  like  a  promissory  note,  may  be  payable  "to  order," 
or  "  bearer,"  and  is  subject  to  protest  in  case  the  payment  or  accept- 
ance is  refused. 

5.  A  time  draft  is  universally  entitled  to   the  three  days  grace: 
but,  in  about  half  of  the  states,  no  grace  is  allowed  on  sight  drafts. 

14.  When  a  bank  loans  money,  it  discounts  the  time,  notes, 


BANK  DISCO  UNT.  271 

drafts,  etc.,  offered  by  the  borrower  at  a  rate  of  discount 
agreed  upon,  but  not  in  accordance  with  the  principles  of 
true  discount. 

15.  Bank  Discount  is   simple   interest  on  the  face  of  a 
note,  calculated  from   the   day  of  discount  to   the   day   of 
maturity,  and  paid  in  advance. 

16.  The  Proceeds  of  a  note  is  the  amount  which  remains 
after  deducting  the  discount  from  the  face. 

REMARKS. — 1.  Since  the  face  of  every  note  is  a  debt  due  at  a  future 
time,  its  cost  ought  to  be  the  present  worth  of  that  debt,  and  the 
bank  discount  should  be  the  same  as  the  true  discount.  As  it  is, 
the  former  is  greater  than  the  latter ;  for,  bank  discount  is  interest  on 
the  face  of  the  note,  while  true  discount  is  the  interest  on  the  present  worth, 
which  is  always  less  than  the  face.  Hence,  their  difference  is  the  interest 
of  the  difference  between  the  present  worth  and  face ;  that  is,  the 
interest  of  the  true  discount.  (See  Art.  310,  3.) 

2.  In  discounting  notes,  the  three  days  of  grace  are  always  taken 
into  account;  and  in  Delaware,  the  District  of  Columbia,  Maryland, 
Missouri,  and  Pennsylvania,  the  day  of  discount  and  the  day  of  matu- 
rity are  both  included  in  the  time. 

3.  If  the   borrower    is    not   the  maker   of   the  notes    or    drafts, 
he  must  be  the  last  indorser,  or  holder  of  them. 

314.  Problems  in  bank  discount  are  solved  like  those  of 
simple  interest.     The   face  of  the  note   corresponds  to  the 
Principal;  the  bank  discount,  to  the  Interest;  the  proceeds, 
to  the  Principal  less  the  Interest;  and  the  term  of  discount, 
to  the  Time. 

CASE    I. 

315.  Given  the  face  of  the  note,  the  rate,  and  time, 
to  find  the  discount  and  proceeds. 

PROBLEM. — What  is  the  bank  discount  of  $770  for  90 
days,  at 


272  RAY'S  HIGHER  ARITHMETIC. 

OPERATION. 

Int.  of  $  1  for  9  3  da.,  at  6^,  —  $.0155  —  Eate  of  Discount. 
$770X-0155  =  $11.935,  Discount. 
$770  —  $11.935  =  $758.065,  Proceeds. 

SOLUTION. — Find  the  interest  of  $770  for  93  da.,  at  6^;  this  is 
$11.935,  which  is  the  bank  discount.  The  proceeds  is  the  difference 
between  $770  and  $11.935,  or  $758.065 

PRINCIPLES. — 1.  The  interest  on  the  sum  discounted  for  the 
given  time  (plus  the  three  days  of  grace)  at  the  given  rate  per 
cent,  is  the  bank  discount. 

2.  The  proceeds  is  equal  to  the  sum  discounted  minus  the 
discount. 

Rule.— 1*  Find  the  interest  on  the  sum  discounted  for  three 
days  more  than  the  given  time,  at  the  given  rate;  it  is  the 
discount. 

2.  Subtract  the  discount  from  the  sum  discounted,  and  the 
remainder  is  the  proceeds. 

EEMARKS. — 1.  As  with  promissory  notes,  the  three  days  of  grace 
are  not  counted  on  a  draft  bearing  the  words  "  without  grace." 

2.  The  following  problems  present,  each,  two  dates, — one  showing 
when  the  note  is  nominally  due,  and  the  other  when  it  is  legally  due. 
Thus,  in  an  example  which  reads,  tl  Due  May  3/6,"  the  first  is  the 
nominal,  and  the  second  the  legal  date. 


EXAMPLES  FOR  PRACTICE. 

Find  the  day  of  maturity,  the  time  to  run,  and  the  pro- 
ceeds of  the  following  notes : 

1.    $792.50  QUINCY,  ILL.,  Jan.  3,  1870. 

Six  months  after  date,  I  promise  to  pay  to  the  order 
of  Jones  Brothers  seven  hundred  and  ninety-two  -ffa  dollars, 
at  the  First  National  Bank,  value  received. 

Discounted  Feb.  18,  at  Gfi.  ALBERT    L.    TODD. 

Due  July  3/6 ;  138  da.  to  run;  proceeds,  $774.27 


BANK  DISCOUNT.  273 

2.   $1962T\50-  NEW  YORK,  July  26,  1879. 

Value  received,  four  months  after  date,  I  promise  to 
pay  B.  Thorns,  or  order,  one  thousand  nine  hundred  sixty- 
two  ^Q  dollars,  at  the  Chemical  National  Bank. 


Discounted  Aug.  26,  at  7  ft.  E<    WlLLIAMS- 

Due  Nov.  26/29;  95  da-  to  run;  pro.  $1926.70 

3.  $2672T^.  PHILADELPHIA,  March  10,  1872. 
Nine  months  after  date,  for  value   received,  I  promise 

to  pay  Edward  H.  King,  or  order,  two  thousand  six  hun- 
dred seventy-two  y1^-  dollars,  without  defalcation. 

JEREMIAH  BARTON. 
Discounted  July  19,  at  6«fo. 

Due  Dec.  10/13;   148  da.  to  run;  pro.  $2606.27 

4.  $3886.  ST.  Louis,  Jan.  31,  1879. 
One  month  after  date,  we  jointly  and  severally  promise 

to  pay  C.  McKnight,  or  order,  three  thousand  eight  hun- 
dred eighty-six  dollars,  value  received,  negotiable  and  pay- 
able, and  without  defalcation  or  discount. 

T.  MONROE, 

I.  FOSTER. 
Discounted  Jan.  31,  at  \\%  a  month. 

Due  Feb.   2^  Mar.;   32  da.  to  run;   pro.  $3823.82 

REMARK.  —  A  note,  drawn  by  two  or  more  persons,  "  jointly  and 
severally,"  may  be  collected  of  either  of  the  makers,  but  if  the  words 
"  jointly  and  severally"  are  not  used,  it  can  only  be  collected  of  the 
makers  as  a  firm  or  company. 

5.  $2850.  AUSTIN,  TEX.,  April  11,  1879. 
For  value  received,  eight  months  after  date,  we  promise 

to  pay  Henry  Hopper,  or  order,  twenty-eight  hundred  and 
fifty  dollars,  with  interest  from  date,  at  six  per  cent  per 

annum.  TT  p    ^ 

HANNA  &  TUTTLE. 

Discounted  June  15,  at  6^. 

Due  Dec.  ll/J4;  182  da.  to  run;  pro.  $2875.47 


274  RA  Y'S  HIGHER  ARITHMETIC. 

6.  $737T4oV  BOSTON,  Feb.  14,  1880. 
Value  received,  two  months  after  date,  I  promise  to  pay 

to  J.  K.  Eaton,   or  order,  seven  hundred   thirty-seven  T40°(j 
dollars,  at  the  Suffolk  National  Bank. 

WILLIAM  ALLEN. 
Discounted  Feb.  23,  at  10^. 

Due  April  14/17;  54  da.  to  run;  pro.  $726.34 

7.  $4085T2^V  NEW  ORLEANS,  Nov.  20,  1875. 
Value  received,  six  months  after  date,  I  promise  to  pay 

John   A.    Westcott,  or  order,  four  thousand  eighty-five  T2^ 
dollars,  at  the  Planters'  National  Bank. 

E.  WATERMAN. 

Discounted  Dec.  31,  1875,  at  5^,. 

Due  May         g,  1876;  144  da.  to  run;  pro.  $4003.50 


CASE     II. 

316.  Given  the  proceeds,  time,  and  the  rate  of  dis- 
count, to  find  the  face  of  the  note. 

PROBLEM.  —  For  what  sum  must  a  60  da.  note  be  drawn, 
to  yield  $1000,  when  discounted,  at  6%  per  annum? 

OPERATION. 

The  bank  discount  of  $1  for  63  da.,  at  6  ft  =  $.0105 
Proceeds  of   $1  —  $.9  895;  $1  00  0  -f-  .  98  9  5  =  $  1  0  1  0  .  6  1,  the 
face  of  the  note. 

SOLUTION.  —  For  every  $1  in  the  face  of  the  note,  the  proceeds,  by 
Case  I,  is  $.9895  ;  hence,  there  must  be  as  many  dollars  in  the  face 
as  this  sum,  $.9895,  is  contained  times  in  the  given  proceeds,  $1000; 
this  gives  $1010.61  for  the  face  of  the  note. 

Rule.  —  Divide  the  given  proceeds  by  the  proceeds  of  $1  for 
the  given  time  and  rate;  or,  divide  the  given  discount  by  the 
discount  of  $1  for  the  given  time  and  rate;  the  quotient  i* 
equal  to  the  face  of  the  note. 


BANK  DISCOUNT.  275 


EXAMPLES  FOR  PKACTICE. 

1.  Find  the   face  of  a  30  da.  note,  which  yields  $1650, 
when  discounted  at  \\cfc  a  mo.  $1677.68 

2.  The  face  of  a  60  da.    note,  which,  discounted  at  6% 
per  annum,  will  yield  $800.  $808.49 

o.  The  face  of  a  90  da.  note,  bought  for  $22.75  less  than 
its  face,  discounted  at  1%.  $1258.06 

4.  The  face  of  a  4  mo.  note,  which,  discounted  at  1  %  a 
month,  yields  $3375.  $3519.29 

5.  The  face  of  a  6  mo.  note,  which,  discounted  at   10^ 
per  annum,  yields  $4850.  $5109.75 

6.  The  face  of  a  60  da.  note,  discounted  at  2%  a  month, 
to  pay  $768.25  $801.93 

7.  The  face  of  a  40  da.  note,  which,  discounted  at  8^, 
yields  $2072.60  .  $2092.60 

8.  The  face   of  a  30  da.   and  90  da.  note,  to  net  $1000 
when  discounted  at  6%. 

$1005.53  at  30  da.;  $1015.74  at  90  da. 

CASE   III. 

317.     Given  the  rate    of  bank   discount,  to  find  the 
corresponding  rate  of  interest. 

PROBLEM. — What  is  the   rate  of  interest  when  a  60  da. 
note  is  discounted  at  2%  a  month? 

OPERATION. 

Assume  $  1  0  0  as  the  face  of  the  note. 
Then,     6  3  days  =  time. 

$4.20=  discount. 
$95.80—  proceeds. 

$.00175=:  interest  of  $  1,  at  1  ft,  for  the  given  time. 
And  $95. 80X- 00175  =  $. 1676  5,  interest  of  proceeds,  at  1  fa 
for  the  given  time ;  then, 

$  4  .  2  0  -f-  $  .  1  6  7  6  5  =  2  5  ^,  rate. 


276  RAY'S  HIGHER  ARITHMETIC. 

SOLUTION. — The  discount  of  $100  for  63  days,  at  2fi  a  month,  is 
$4.20,  and  the  proceeds  is  $95.80  The  interest  of  $1  for  the  given 
time,  at  If0l  is  $.00175,  and  of  $95.80  is  $.16765  To  find  the  rate, 
we  proceed  thus  :  $4.20  -f-  $.16765  =  25¥2/¥. 

Rule. — 1.  Find  the  discount  and  proceeds  of  $100  or  81 
for  the  time  the  note  runs. 

2.  Divide  the  discount  by  the  interest  of  the  proceeds,  at  1^, 
for  the  same  time;  the  quotient  is  the  rate. 


EXAMPLES  FOR  PRACTICE. 

What  is  the  rate  of  interest : 

1.  AVhen  a  30  da.  note  is  discounted  at  1,  1^,  1J,  2%  a 
month?  12-J.ff,  15^,  18T^4T,  24^%  per  annum. 

2.  When  a   60  da.  note  is  discounted  at  6,  8,  10^   per 
annum ?  6iV%>  $  AV  lO^y^  per  annum. 

3.  When   a   90   da.   note   is   discounted  at   2,  21,  3%  a 
month  ?  25f£f ,  32T%%,  39f ffify  per  annum. 

4.  When  a  note  running  1  yr.  without  grace, is  discounted 
at  5, 6, 7, 8, 9, 10,  \2%  ?   _5^,  6tf ,  7#,  8tf,  9ft,  111  IB&%. 

5.  My  note,  which   will  be  legally  due  in  1  yr.  4  mo.  20 
da.,  is  discounted  by  a  banker,  at  8%:  what  rate  of  interest 
does  he  receive?  9^. 

REMARK. — It  may  seem  unnecessary  to  regard  the  time  the  note 
has  to  run,  in  determining  the  rate  of  interest;  but,  a  comparison 
of  examples  1  and  3,  shows  that  a  90  da.  note,  discounted  at  2<fa  a 
month,  yields  a  higher  rate  of  interest  than  a  30  da.  note  of  the  same 
face,  discounted  at  2 ft  a  month.  The  discount,  at  the  same  rate,  on 
all  notes  of  the  same  face,  varies  as  the  time  to  run,  and  if  in  each 
case,  it  was  referred  to  the  same  principal,  the  rate  of  interest  would 
be  the  same ;  but  when  the  discount  becomes  larger,  the  proceeds  or 
principal  to  which  it  is  referred,  becomes  smaller,  and  therefore  the 
rate  of  interest  corresponding  to  any  rate  of  discount  increases  with  the  time 
the  note  has  to  run.  Hence,  the  profit  of  the  discounter  is  greater 
proportionally  on  long  notes  than  on  short  ones,  at  the  same  rate. 


BANK  DISCOUNT.  277 


CASE   IV. 

318.  Given  the  rate  of  interest,  to  find  the  corre- 
sponding rate  of  discount. 

PROBLEM. — What  is  the  rate  of  discount  when  a  60  day 
note  yields  2%  interest  a  month? 

OPERATION. 

Assume  $  1  0  0  as  the  face  of  the  note. 
Then,  6  3  days  =  the  time  ; 

$  4 .  2  0  —  the  interest ; 
and  $104.20  =  the  amount ; 
also,  $.18235  =  interest  of  amount,  at  1  ^,  for  the  given  time; 

$4.  20-*-$.  18235  =  23  ^7r,  rate  per  cent. 

REMARK. — Note  the  difference  between  this  solution  and  that 
under  the  preceding  case.  In  that,  the  interest  was  found  on  the 
proceeds ;  in  this,  it  is  computed  on  the  amount. 

Rule.- — 1.  Find  the  interest  and  the  amount  of  $100  or  $1 
for  the  given  time. 

2.  Divide  the  interest  by  the  interest  of  the  amount  at  1% 
for  the  given  time;  the  quotient  is  the  corresponding  rate  of 
discount. 


EXAMPLES  FOR  PRACTICE. 

What  rates  of  discount  : 

1.  On  30  da.  notes,  yield  10,  15,  20^  interest? 


2.  On  60  da.  notes,  yield  6,  8,  10%  interest? 

^1895      74 
°2"ofT>     '5" 

3.  On  90  da.  notes,  yield  1,  2,  4%  a  month  interest? 


4.  On  notes  running  1  yr.  without  grace,  yield  5,  6,  7,  8, 
9,  10^  interest?  4Jf  5ff, 


278  RAY'S  HIGHER  ARITHMETIC. 

V.    EXCHANGE. 
DEFINITIONS. 

319.  1.  Exchange  is  the  method  of  paying  a  debt  in  a 
distant  place  by  the  transfer  of  a  credit. 

REMARK. — This  method  of  transacting  business  is  adopted  as  a 
convenience :  it  avoids  the  danger  and  expense  of  sending  the 
money  itself. 

2.  The  payment  is  effected  by  means  of  a  Bill  of  Exchange. 

3.  A  Bill  of  Exchange   is  a  written  order  on  a  person 
or  company  in  a  distant  place  for  the  payment  of  money. 

REMARK. — The  term  includes  both  drafts  and  checks. 

4.  Exchange  is  of  two  kinds :    Domestic,  or  Inland,  and 
Foreign. 

5.  Domestic  Exchange  treats  of  drafts  payable  in  the 
country  where  they  are  made.     (See  page  269  for  form.) 

6.  Foreign    Exchange    treats    of    drafts    made   in    one 
country  and  payable  in  another. 

7.  A  foreign   bill  of  exchange  is  usually  drawn  in  trip- 
licate,  called   a  Set   of  Exchange;    the    different   copies, 
termed  respectively  the  first,  second,  and  third  of  exchange, 
are  then  sent  by  different  mails,  that  miscarriage  or  delay 
may  be  avoided.     When   one  is  paid,   the   others  are  void. 
The  following  is  a  common  form : 

FOREIGN  BILL  OF  EXCHANGE. 
1.  Exchange  for  NEW  YORK,  July  2,  1878. 

£1000.  Thirty  days    after  sight  of  this  First  of 

Exchange  (Second  and  Third  of  same  tenor  and  date  unpaid), 
pay  to  the  order  of  James  S.  Rollins  One  Thousand  Pounds 
Sterling,  value  received,  and  charge  to  account  of 

To  JOHN  BROWN  &  Co.,  J.  S.  CHICK. 

No.  1250.  LIVERPOOL,  ENGLAND. 


EXCHANGE.  279 

8.  The  Rate  of  Exchange  is  a  rate  per  cent  of  the  face 
of  the  draft. 

9.  The  Course  of  Exchange  is  the  current    price    paid 
in  one  place  for  bills  of  exchange  on  another. 

10.  The  Par  of  Exchange  is  the  estimated  value  of  the 
monetary  unit  of  one  country  compared  with  that  of  another 
country,  and  is  either  Intrinsic  or  Commercial. 

11.  Intrinsic  Par  of  Exchange   is  based   on   the   com- 
parative weight  and  purity   of  coins. 

12.  Commercial    Par  of  Exchange   is  based   on  com- 
mercial usage,  or  the  price  of  coins  in  different  countries. 

DOMESTIC  EXCHANGE. 

320.  Where  time  is  involved,  problems  in  Domestic 
Exchange  are  solved  in  accordance  with  the  principles  of 
Bank  Discount. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  paid  for  $3805.40   sight  exchange  on  Boston, 
at  \%  premium?  $3824.43 

2.  What  is  the  cost  of  a  check  on  St.  Louis  for  $1505.40, 
at  \<fa   discount?  $1501.64 

3.  What  must  be  the  face  of  a  sight  draft  to  cost  $2000, 
at  f  ^  premium?  $1987.58 

4.  What  must  be  the  face  of  a  sight  draft  to  cost  $4681.25, 
at  \\%  discount?  $4740.51 

5.  What  will  a  30  day  draft  on  New  Orleans  for  $7216.85 
cost,  at  \%  discount,  interest  6%  ? 

OPERATION. 

$1  —  $.00f  =  $.99625,  rate  of  exchange. 

$.0055     —  bank  discount  of  $  1  for  3 3  da. 
$.99075  =  cost  of  exchange  for  $  1. 
$7216.  85X.99075  =  $7150.  094  +,  Ans. 


280  RAY'S  HIGHER  ARITHMETIC. 

SOLUTION. — From  the  rate  of  exchange  subtract  the  bank  dis- 
count of  $1  for  33  days,  at  6^;  the  remainder  is  the  cost  of  exchange 
for  $1.  Then  multiply  the  face  of  the  draft  by  the  cost  of  exchange 
for  $1,  which  gives  $7150.094,  the  cost  of  the  draft. 

6.  What  will  a  60  day  draft  on  New  York  for  $12692.50 
cost,  at  |^  premium,  interest  6%?  $12654.42 

7.  What   must   be  the   face   of  an  18  day  draft,  costing 
$5264.15,  at  \%  premium,  interest  6^  ?  $5256.27 

8.  What   must   be    the   face   of  a   21    day  draft,  costing 
$6836.75,  at  \%  discount,  interest  6%  ?  $6925.04 

9.  A  commission  merchant  in  St.  Louis  sold  5560  Ib.  of 
bacon,   at  11^  ct.  a  Ib.;    his   commission  is   2^^,  and   the 
course  of  exchange  98J^£:  what  is  the  amount  of  the  draft 
that  he  remits  to  his  consignor?  $632.91 

10.  A  grain  merchant  in  Toledo  sold  11875  bu.  of  corn, 
at  40  cents  a  bushel;  deducting  3%  as  commission,  he  pur- 
chased a  60  day  draft  with  the  proceeds,  at  2%  premium; 
required  the  face  of  the  draft?  $4564.14 

11.  Sold   lumber  to  the   amount  of  $20312.50,  charging 
\\%  commission,  and  remitted  the  proceeds  to  my  consignor 
by  draft;  required  the  face  of  the  draft,  exchange  \%  dis- 
count? $20108.35 

12.  If  a  45  day  draft  for  $5500  costs  $5538.50,  find  the 
rate  of  exchange. 

OPERATION. 

The  bank  discount  of  $5500  for  48  days,  at  6  ^=$4 4. 
Then  $5538  .50 +  $  44  ,-$5500  =$8  2.  5  0  premium; 

$82.50 
And  |5500   =  •  0  1  5  =  1  \  fa  rate  of  exchange. 

13.  A  father  sent  to  his  son,  at  school,  a  draft  for  $250, 
at  3  mo.,  interest  6%,  paying  $244.25  for  it;  find  the  rate 
of  exchange.  f  %  discount. 

14.  An    agent    owing    his  principal    $1011.84,    bought   a 
draft  with  this  sum  and  remitted  it ;  the  principal  received 
$992  ;  find  the  rate  of  exchange.  2^  premium. 


EXCHANGE. 


281 


FOKEIGN   EXCHANGE. 

321.  1.  In  Foreign  Exchange  it  is  necessary  to  find  the 
value  of  money  in  one  country  in  terms  of  the  monetary  unit 
of  another  country.  We  reduce  foreign  coins  by  comparison 
with  U.  S.  money,  to  find  their  value. 

KEMARK.-  By  an  Act  of  Congress,  March  3,  1873,  the  Director 
of  the  Mint  -is  authorized  to  estimate  annually  the  values  of  the 
standard  coins,  in  circulation,  of  the  various  nations  of  the  world. 
In  compliance  with  this  law,  the  Secretary  of  the  Treasury  issued 
the  following  estimate  of  values  of  foreign  coins,  January  1,  1884: 


COUNTRY. 

MONETARY   UNIT. 

VALUE   IN 
U.  S.  MONEY. 

Austria  

Florin  

$   398 

Bolivia  •••          .    ..      

806 

Brazil         .        .                

Milreis  of   1000  reis... 

546 

Chili   

Peso  

912 

Cuba  ..      

Peso  

932 

Denmark  Norway  Sweden  ... 

Crown   

268 

"Ro'VDt 

Piaster  

049 

-L/t3./Ft  •• 
France   Belgium    Switz  • 

Franc  

193 

Great  Britain  

Pound   sterling  

4.866J 

Mark     

.238 

Rupee  of  16  annas-.  ••••• 

.383 

Japan  

Yen  (silver)  

.869 

Dollar  

875 

Netherlands 

Florin 

402 

Milreis  of  1000  reis... 

1  08 

Rouble  of  100  copecks 

.645 

Tripoli  

Miahbub  of  20  piasters 

.727 

Turkey  

Piaster  »     

.044 

NOTE. — The  Drachma  of  Greece,  the  Lira  of  Italy,  the  Peseta  (100 
centimes)  of  Spain,  and  the  Bolivar  of  Venezuela,  are  of  the  same 
value  as  the  Franc.  The  Dollar,  of  the  same  value  as  our  own,  is  the 
standard  in  the  British  Possessions,  N.  A.,  Liberia,  and  the  Sandwich 
Islands.  The  Peso  of  Ecuador  and  the  United  States  of  Colombia, 

and  the  Sol  of  Peru,  are  the  same  in  value  as  the  Boliviano. 
H.  A.  24. 


282  RA  Y'S  HIGHER  ARITHMETIC. 

2.  Bills  of  Exchange  on   England,  Ireland,  and  Scotland 
are  bought  and  sold  without  reference  to  the  par  of  exchange. 

3.  It   is   customary  in  foreign  exchange  to  deal   in  drafts 
on  the  various  commercial  centers.     London  exchange  is  the 
most  common,   and  is  received  in  almost  all  parts  of  the 
civilized  world. 

BEMARK. — London  exchange  is  quoted  in  the  newspapers  in 
several  grades:  as,  "Prime  bankers'  sterling  bills,"  or  bills  upon 
banks  of  the  highest  standing ;  "  good  bankers,"  next  in  rank ; 
"  prime  "  and  "  good  commercial,"  or  bills  on  merchants,  etc.  The 
prices  quoted  depend  upon  the  standing  of  the  drawee  and  upon 
the  demand  for  the  several  classes  of  bills.  Usually  a  double  quo- 
tation, one  for  60  day  bills,  and  the  other  for  3  day  bills,  is  given 
for  each  class. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  cost  of  a  bill  of  exchange  on  London,  at  3 
days  sight,  for  £530  12s.,  exchange  being  quoted  at  $4.88 

OPERATION. 

£530   12s.  =  £530.  6 

$4.  88X530. 6  =  $2589.  328 

2.  Find  the  cost  of  a   bill   of  exchange  on  London,  at 
sight,   for  £625  10s.   10d.,  when  exchange  is  $4.87  pound 
sterling.  $3046.39 

3.  What  is   the  cost  of  a  bill  on  Paris  for  1485  francs, 
$1  —  5.15  francs.  $288.35 

4.  What  is  the  cost   of  a  bill  on   Amsterdam  for  4800 
guilders,  quoted  at  41J  cents,  brokerage  \<fa  ? 

OPERATION. 

$ .  4  1  5  X  4  8  0  0  =  $  1  9  9  2  . 
$1992X.OOi=r$9.96,  brokerage. 
$1992-f-$9.96  =  $2001.96,  cost  of  bill. 

5.  What  will  a  draft  in  St.  Petersburg  on  New  York  for 
$5000  cost,  if  a  rouble  be  worth  $.74?  6756ff  roubles. 


ARBITRATION  OF  EXCHANGE.  283 

6.  What  will  a  draft  on  Charleston  for  $4500  cost  at  Rio 
Janeiro  (milreis  =$.54),  discount  2%?         8166. 66f  milreis. 

7.  A  gentleman  sold  a  60  day  draft,  which  was  drawn  on 
Amsterdam  for  1000  guilders;  he  discounted,  it  at  6%,  and 
brokerage  was  \%'.  what  did  he  get  for  it,  a  guilder  being 
valued  at  40£  cents?  $397.77  -f 


AKBITRATION  OF  EXCHANGE. 
DEFINITIONS. 

322.  1.  Owing  to  the  constant  variation  of  exchange,  it 
is  sometimes  advantageous  to  draw  through  an  intermediate 
point,  or  points,  in  place  of  drawing  directly.  This  is  called 
Circular  Exchange. 

2.  The  process  of  finding  the  proportional  exchange  be- 
tween two  places  by  means  of  Circular  Exchange,  is  called 
the  Arbitration  of  Exchange. 

3.  Simple   Arbitration   is    finding  the   proportional  ex- 
change when  there  is  but  one  intermediate  point. 

4.  Compound    Arbitration    is    finding  the   proportional 
exchange  through  two  or  more  intermediate  points. 

PROBLEM.  —A  merchant  wishes  to  remit  $2240  to  Lisbon : 
is  it  more  profitable  to  buy  a  bill  directly  on  Lisbon,  at  1 
milreis  =  $1.10 ;  or  to  remit  through  London  and  Paris,  at 
£1  =  84.88  =  25.20  francs,  to  Lisbon,  1  milreis  =  5.5  francs? 

OPERATION. 

$2240-^-1.10  —  2036.364  milreis,  direct  exchange. 

(  )  milreis  =$2  2  40. 
$4.88  =  25.20  francs. 
5  . 5  francs  =  1  milreis. 

2240X25. 20X  1 


:=2103.129  milreis,  circular  exchange. 
4 .  8  8  X  5  .  5 

Hence,  the  gain  is  6  6  . 7  6  5  milreis  in  buying  circular  exchange. 


284  RA  Y'S  HIGHER  ARITHMETIC. 

SOLUTION.--  -By  direct  exchange  $2240  will  buy  2036.364  milreis. 
By  circular  exchange  we  have  a  set  of  equations,  all  of  whose 
members  are  known  except  one.  Multiplying  the  right-hand 
members  together  for  a  dividend,  and  dividing  the  product  by  the 
product  of  the  left-hand  members,  the  quotient  is  the  required  mem- 
ber, which,  in  this  case,  is  2103.129  milreis.  The  difference  in  favor 
of  the  circular  method  is  66.765  milreis. 

PROBLEM.  —  A  merchant  in  Memphis  wishes  to  remit 
$8400  to  New  York;  exchange  on  Chicago  is  \\%  premium, 
between  Chicago  and  Detroit  1%  discount,  and  between 
Detroit  and  New  York  \%  discount:  what  is  the  value  of 
the  draft  in  New  York,  if  sent  through  Chicago  and  Detroit? 

OPERATION. 

($        )  N.  Y.  ==  $  8  4  0  0,  Memphis. 
$  1  .  0  1  \  Mem.  =  $  1,  Chicago. 
$.99  Chi.  =  $1,  Detroit. 
$  .  9  9  J  Det.  =  $  1,  N.  Y. 

^400X1X1X1 


1.015X-99X-995 

SOLUTION.  —  By  the  problem,  $1.015  in  Memphis  =$1  in  Chicago; 
$.99  in  Chicago  =  $1  in  Detroit  ;  and  $.995  in  Detroit  =  $1  in  New 
York.  Hence,  multiplying  and  dividing,  as  in  the  preceding 
problem,  we  obtain  the  answer,  $8401.46 

Rule  1.  —  1.  Form  a  series  of  equations,  expressing  the  con- 
ditions of  the  question;  the  first  containing  the  quantity  given 
equal  to  an  unknown  number  of  the  quantity  required,  and  all 
arranged  in  such  a  way  that  the  right-hand  quantity  of  each 
equation  and  the  left-hand  quantity  in  the  equation  next  follow- 
ing, shall  be  of  the  same  denomination,  and  also  the  right-hand 
quantity  of  the  last  and  the  left-hand  quantity  of  the  first. 

2.  Cancel  equal  factors  on  opposite  sides,  and  divide  the 
product  of  the  quantities  in  the  column  which  is  complete  by 
the  product  of  those  in  the  other  column.  The  quotient  will  be 
the  quantity  required. 


ARBITRATION  OF  EXCHANGE.  285 

Rule  2. — Reduce  the  given  quantity  to  the  denomination 
with  which  it  is  compared',  reduce  this  result  to  the  denomina- 
tion with  which  it  is  compared;  and  so  on,  until  the  required 
denomination  is  reached,  indicating  the  operations  by  writing 
as  multipliers  the  proper  unit  values.  The  compound  fraction 
thus  obtained,  reduced  to  its  simplest  form,  uill  be  the  amount 
of  the  required  denomination. 

NOTES. — 1.  The  first  is  sometimes  called  the  chain  rule,  because 
each  equation  and  the  one  following,  as  well  as  the  last  and  first, 
are  connected  as  in  a  chain,  having  the  right-hand  quantity  in  one 
of  the  same  denomination  as  the  left-hand  quantity  of  the  next. 

2.  In   applying  this  rule  to  exchange,  if  any  currency  is   at  a 
premium  or  a  discount,  its   unit  value  in  the  currency  with  which 
it  is  compared,  must  be  multiplied  by  such  a  mixed  number,  or 
decimal,  as  will  increase  or  diminish  it  accordingly. 

3.  If  commission  or  brokerage  is  charged  at  any  point  of  the 
circuit,  for   effecting  the  exchange  on  the  next  point,  the  sum  to  be 
transmitted  must  be  diminished  accordingly,  by  multiplying  it  by 
the  proper  number  of  decimal  hundredths. 


EXAMPLES  FOR  PRACTICE. 

1.  A,  of  Galveston,  has  $6000  to  pay  in  New  York.     The 
direct  exchange  is   \^0    premium;    but   exchange  on    New 
Orleans  is  \%   premium,  and   from  New  Orleans  to  New 
York   is  \%   discount.     By   circular   exchange,  how  much 
will  pay  his  debt,  and  what  is  his  gain  ? 

$5999.96;  830.04  gain. 

2.  A  merchant  of  St.  Louis  wishes  to  remit  $7165.80  to 
Baltimore.     Exchange  on  Baltimore  is  \%   premium ;  but, 
on  New  Orleans  it  is  \%  premium  ;  from   New  Orleans  to 
Havana,   \%    discount;    from   Havana    to   Baltimore,   \^0 
discount.      What  will   be   the  value  in  Baltimore   by  each 
method,  and  how  much  better  is  the  circular? 

Direct,  $7147.93;  cir.,  $7183.77;  gain,  $35.84 


286  RAY'S  HIGHER  ARITHMETIC. 

3.  A  Louisville  merchant  has  f  10000  due  him  in  Charles- 
ton.    Exchange  on  Charleston  is  \%  premium.     Instead  of 
drawing  directly,  he  advises  his  debtor  to  remit  to  his  agent 
in   New  York  at  \%    premium,  on  whom  he  immediately 
draws  at  12   da.,  and  sells  the  bill  at  \%  premium,  interest 
off  at  6^.     What  does  he  realize,  and  what  gain  over  the 
direct  exchange?  Realizes  $10029.94;  gain,  $17.44 

SUGGESTION. — Interest  at  6^  per  annum  is  \fi  a  month,  or  \<J0 
for  15  days,  which  diminishes  the  rate  of  premium  to  \Gjc. 

4.  A   Cincinnati   manufacturer   receives,   April   18th,   an 
account  of  sales  from  New  Orleans;  net  proceeds  $5284.67, 
due  June  4/?»     He  advises  his  agent  to  discount  the  debt  at 
6%,  and  invest  the  proceeds  in  a  7  day  bill  on  New  York,  in- 
terest off  at  6  % ,  at  \^0  discount,  and  remit  it  to  Cincinnati. 
The  agent  does  this,  April  26.     The  bill  reaches  Cincinnati 
May  3,  and  is  sold  at  \%  premium.     What  is  the  proceeds, 
and  how  much  greater  than  if  a  bill  had  been  drawn  May  3, 
on  New  Orleans,  due  June  7,  sold  at  \^0    premium,  and 
interest  off  at  6%  ?  Proceeds,  $5296.10;  gain,  $35.65 

5.  A  merchant  in  Louisville  wishes  to  pay  $10000,  which 
he  owes  in  Berlin.     He  can  buy  a  bill  of  exchange  in  Louis- 
ville on  Berlin  at  the  rate  of  $.96  for  4  reichmarks;  or  he  is 
offered  a  circular  bill  through  London  and  Paris,  brokerage 
at  each   place  \%,  at  the  following  rates:  £1  =  $4.90  = 
25.38   francs,   and  5  francs  =  4  reichmarks.     What  is  the 
difference  in  the  cost?  $81.35 


VI.    EQUATION  OF   PAYMENTS. 
DEFINITIONS. 

323.  1.  Equation  of  Payments  is  the  process  of  find- 
ing the  time  when  two  or  more  debts,  due  at  different  times, 
may  be  paid  without  loss  to  either  debtor  or  creditor. 


EQ  UA  TION  OF  PA  YMENTS.  287 

2.  The  time  of  payment  is  called  the  Equated  Time. 

3.  The  Term  of   Credit  is   the  time  to  elapse  before  a 
debt  is  due. 

4.  The  Average  Term  of   Credit  is  the  time  to  elapse 
before  several    debts,  due  at   different   dates,   may  be  paid 
at  once. 

5.  To    Average    an  Account   is    to   find    the   equitable 
time  of  payment  of  the  balance. 

6.  Settling  or  Closing  an  Account  is  finding  how  much 
is  due  between  the  parties  at  a  particular  time.     It  is  some- 
times called  Striking  a  Balance. 

7.  A  Focal   Date   is   a  date    from   which   we   begin   to 
reckon  in  averaging  an  account. 

324.  Equation  of  Payments  is  based  upon  the  following 
principles : 

PRINCIPLES. — 1.  That  any  sum  of  money  paid  before  it  is 
due,  is  balanced  by  keeping  an  equal  sum  of  money  for  an 
equal  time  after  it  is  due. 

2.  That  the  interest  is  the  measure  for  the  use  of  any  sum 
of  money  for  any  time. 

REMARK. — The  first  statement  is  not  strictly  correct,  although  it 
has  the  sanction  of  able  writers,  and  is  very  generally  accepted. 
For  short  periods,  it  will  make  no  appreciable  difference. 

PROBLEM. — A  buys  an  invoice  of  $250  on  3  mo.  credit; 
$800  on  2  mo.;  and  $1000  on  4  mo.;  and  gives  his  note  for 
the  whole  amount :  how  long  should  the  note  run  ? 

OPERATION. 
Debts.      Terms.      Equivalents. 

250     X     3     =     750 

800     X     2     =  1600 
1000     X     4    =  4000 


2050  6350 

time  to  run  = —  mo.  —  3  mo.  3  da.,  Ans. 

2  0  o  0 


288  AM  F'tf  HIGHER  ARITHMETIC. 

SOLUTION. — The  use  of  $1  for  6350  months  is  balanced  by  the  use 
of  $2050  for  3  months  and  3  days.  Or,  the  interest  of  $1  for  6350 
months  equals  the  interest  of  $2050,  3  months  3  days,  at  the  same 
rate. 

PROBLEM. — I  buy  of  a  wholesale  dealer,  at  3  mo.  credit, 
as  follows:  Jan.  7,  an  invoice  of  $600;  Jan.  11,  $240;  Jan. 
13,  $400;  Jan.  20,  $320;  Jan.  28,  $1200;  I  give  a  note  at 
3  mo.  for  the  whole  amount :  when  is  it  dated  ? 

OPERATION. 

$600X     0  =  $  1  for  0  days. 

$240X     4  =  $1    "         960     " 

$400X     6=$1    "      2400     " 

$320X13=$!    "     4160     " 

$1200  X  2  1  —  $  1    "  25200     " 

$2760  $1" 32720" 

Whence    32720-5-2760  =  11.8  +  days ;  hence   the   time   is   1 2 
days,  nearly,  after  Jan.  7 ;  that  is,  Jan.   1  9. 

SOLUTION. — Start  at  the  date  of  the  first  purchase,  and  proceed  as 
in  the  preceding  solution. 

325.  When  the  terms  of  credit  begin  at  the  same  date, 
we  have  the  following  rule : 

Rule. — Multiply  each  debt  by  its  term  of  credit,  and  divide 
the  sum  of  the  products  by  the  sum  of  the  debts;  the  quotient 
ivill  be  the  equated  time. 

326.  If  the   account  has  credits  as  well  as  debits,  it  is 
called   a  compound  equation,  and   may  be    averaged   on  the 
same  principle  as  the  simple  equation. 

The  following  problem  will  illustrate  the  difference  between 
simple  and  compound  equations,  as  it  involves  both  debits 
and  credits : 

PROBLEM. — What  is  the  balance  in  the  following  account, 
and  when  is  it  due? 


EQUATION  OF  PAYMENTS. 


A  in  acc't  with 


289 


DR. 


CR. 


1875. 
Jan. 
Feb. 
Mar. 

14 

1 
10 

To  invoice,  2  mo. 

Da. 
13 
81 

70 
85 
137 

Dol. 
400 
200 
500 
800 
300 

2200 
1575 

625) 

O, 

Prod. 
5200 
6200 
35000 
68000 
41100 

1875. 
Mar. 
Apr. 

May 
June 

1 
15 

20 
4 

By  cash, 
"  remittance 
May  1, 
"  cash, 
"  remittance 
June  25, 

Da. 
0 

61 

80 

116 

Dol. 
500 

250 
375 

450 
1575 

Cf. 

Prod. 
0 

15250 
300CO 

52200 
97450 

May 

25 
16 

155500 
97450 

58050 

93  days  after  March  1  =  June  2. 

SOLUTION. — Start  with  March  1,  the  earliest  day  upon  which  a 
payment  is  made  or  falls  due.  Find  the  products,  both  on  the 
Dr.  and  Cr.  side  of  the  account,  using  the  dates  when  the  items 
are  due,  as  cash.  The  balance  due,  $625,  is  found  by  subtracting 
the  amount  of  the  credits  from  the  amount  of  the  debits.  The 
balance  of  products,  58050,  divided  by  the  balance  of  items,  625, 
determines  the  mean  time  to  be  93  days  after  March  1  =  June  2. 

Rule. — 1.  Find  when  each  item  is  due,  and  take  the  earliest 
or  latest  date  as  the  focal  date. 

2.  Find   the  difference   between   the  focal   date   and  the  re- 
maining  dates,  and   midtiply   each   item    by   its    corresponding 
difference. 

3.  Find  the   difference   between  the  Dr.  and  Cr.  items,  and 
also  between  the  Dr.  and  Cr.  products,  and  divide  the  difference 
of  the  products  by  the  difference  of  the  items.     Add  the  quotient 
to  the  focal  date  if  it  be  the  first  date,  or  subtract  if  it  be  the  last 
date;  the  result  in  either  case  will  be  the  equated  time. 

4.  If  the  two  balances  be  on  opposite  sides  of  the  account,  the 
quotient  must  be  subtracted  from  the  focal  date  if  the  first,  or 
added  if  it  be  the  last  date. 

REMARK. — The  solutions  thus  far  have  been  by  the  product  method. 
The  same  result  will  be  obtained  by  dividing  the  interest  of  the 
product  balance  for  1  day  by  the  interest  of  the  item  balance  for  1 
day.  This  depends  upon  the  principle  that  if  both  dividend  and 
divisor  are  multiplied  or  divided  by  the  same  number  the  quotient 

is  unchanged. 
H.  A.  25. 


290 


RAY'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

What  is  the  mean  time  of  the  following  invoices: 
1.  A  to  B.  DR. 


Dol. 

at. 

Wlien 

due 

Days  after 

Products. 

1877. 

May 

15 

To  invoice  at  4  months. 

800 

June 

1 

'             4 

700 

10 

4 

900 

July 

20 

«             4          * 

600 

Aug. 

1 

4           * 

500 

15 

*              4           * 

1000 

Sum  total, 

4500 

Oct.  30,  1877. 
REMARK. — Start  with  Sept.  15,  the  day  the  first  debt  is  due. 

2.     DR.  E  in  acc't  current  with  F.  CR. 


Feb. 

4 

To  invoice  3  mo. 

$550 

Prod. 

May 

8 

By  cash, 

$150 

Prod. 

Mar. 

'20 

"           "       3    " 

260 

— 

" 

2<i 

"    remit.,  Junes, 

420 

— 

Apr. 

1 

"            "        3    " 

150 

— 

June 

3 

"    note.  1  month, 

340 

— 

n 

5 

.(           «        3    « 

325 

~ 

July 

1 

4*      2       " 

170 

" 

3. 


E  owes  F  $205,  due  Feb.  26. 
H.  Wright  to  Mason  &  Giles.  DR. 


Dol. 

Ct. 

When 

due 

Days  after 

Products. 

1876. 

April'  8. 

Feb. 

1 

To  invoice  3  months. 

900 

20 

"3         " 

700 

Mar? 

10 

"3 

600 

Apr. 
May 

8 
10 

"    cash, 
"       3        " 

500 
900 

June 

15 

«t          g            <( 

400 

Sum  total, 

4000 

Due  June  13,  1876. 

REMARK. — Start  with  April  8,  the  time  the  first  payment  is  due. 
4.     DR.  A  in  acc't  current  with  B.  CR. 


Mar. 

19 

To  invoice  cash, 

$900 

Prod. 

Feb. 

20 

Bv  cash, 

i^OO 

Prod, 

Apr. 

20 

"          *'          " 

HOO 

— 

Mar. 

5 

"    remit.,  Mar.  15, 

300 

— 

June 

15 

<i              **              «i 

700 

— 

June 

2(1 

"    cash, 

200 

— 

May 

10 

600 

July 

10 

500 

~ 

A  owes  B  a  balance  of  $1600,  due  April  23. 


EQUATION  OF  PAYMENTS. 


291 


REMARK.— Start  with  Feb.  20,  and  date  the  remittance,  March  15, 
the  day  it  is  received. 


5.     DR. 


C  in  acc't  current  ivith  D. 


CR. 


Jan. 

4 

To  invoice  2  mo. 

$250 

Prod. 

Mar. 

10 

By  cash, 

£350 

Prod. 

Feb. 

9 

"          "       1    " 

140 

— 

•' 

21 

200 

— 

Apr. 

lf> 

2 

2    •' 
cash, 

450 
100 



Apr. 
May 

4 

20 

"    note,  2  mo. 
"    remit.,  May  25, 

240 
120 

i 

June 

Hi 

"  accep.  16  da.  sight 

500 

C  is  Cr.  $470,  due  Aug.  12. 

6.  I  owe  $912,  due  Oct.  16,  and  $500,  due  Dec.  20.     If 
I  pay  the  first,  Oct.  1,  15  days  before  due,  when   should  I 
pay  the  last?  Jan.  16,  next,  27  days  after  due. 

7.  Oct.  3,  I  had  two   accounts,  amounting  to  $375,  one 
due  Dec.  6,  and  the  other  Nov.  6,  but  equated  for  Nov.  16: 
what  was  each  in  amount?  $250,  $125. 

8.  A  owes  $840,  due  Oct.  3  ;  he  pays  $400,  July  1 ;  $200, 
Aug.  1:  when  will  the  balance  be  due? 

April  30,  next  year. 

9.  I  owe  $3200,  Oct.  25 ;  I  pay  $400,  Sept.   15 ;    $800, 
Sept.  30:  when  should  the  balance  be  paid?  Nov.  12. 

10.  An  account  of  $2500  is  due  Sept.  16 ;  $500  are  paid 
Aug.    1 ;    $500,  Aug.   11 ;  $500,  Aug.  21 :    when  will   the 
balance  be  due?  Nov.  9. 

11.  Exchange  the  five  following  notes  for  six  others,  each 
for  the  same  amount,  and  payable  at  equal  intervals:  one 
of  $1200,  due  in  41  days;  one  of  $1500,  due  in  72  days; 
one  of  $2050,    due  in  80  days;  one  of  $1320,   due  in  110 
days;  one  of  $1730,  due  in  125  days;  total,  $7800. 

The  notes  are  $1300  each,  and  run  25,  50, 
75,  100,  125,  150  days  respectively. 

12.  Burt  owed  in  two  accounts  $487 ;  neither  was  to  draw 
interest  till  after  due, — one  standing  a  year,  and  the  other  two 
years.     He  paid  both  in  1  yr.  5  mo.,  finding  the  true  dis- 
count of  the  second,  at  6%,  exactly  equal  to  the  interest  of 
the  first :  what  difference  of  time  would  the   common  rule 
have  made?  3  days* 


292 


RAY'S  HIGHER  ARITHMETIC. 


VII.    SETTLEMENT  OF  ACCOUNTS. 
DEFINITIONS. 

327.  1.  An  Account  is  a  written  statement  of  debit  and 
credit  transactions   between    two    parties,    giving   the    date, 
quantity,  and  price  of  each  item. 

2.  An  Account  Current  is  a  written   statement  of  the 
business  transactions  between  two  parties,  for  a  definite  time. 

328.  In    settling  an   account,   the   parties   may  wish  to 
find: 

(1).    When  the  balance  is  equitably  due. 

(2).    Wliat  sum,  at  a  given  time,   should  be  paid  to  balance 
the  account. 

The  first  process  is  Averaging  the  Account  (Art.  324)  ;  the 
second  is  Finding  the  Cash  Balance. 


DR. 


Henry  Armor  in  acc't  with  City  Bank. 


CR. 


1875. 

Dol. 

a. 

1874. 

Dol. 

at. 

Jan. 

5 

To  check, 

300 

Dec. 

31 

Bv  balance  old  account, 

500 

" 

20 

•'        " 

500 

1875. 

" 

•a 

tt        « 

100 

Jan. 

7 

"    cash, 

50 

" 

27 

«(                  H 

850 

15 

400 

31 

75 

24 

1000 

EXPLANATION. — The  two  parties  to  this  account  are  Henry 
Armor  and  City  Bank.  The  left-hand,  or  Dr.  side,  shows,  with 
their  dates,  the  sums  paid  by  the  bank  on  the  checks  of  Henry 
Armor,  for  which  he  is  their  debtor.  The  right-hand,  or  Cr.  side, 
shows,  with  their  dates,  the  sums  deposited  in  the  bank  by  Henry 
Armor,  for  which  he  is  their  creditor. 

329.  Generally,  in  an  account  current,  each  item  draws 
interest  from  its  date  to  the  day  of  settlement. 


SETTLEMENT  OF  ACCOUNTS. 


293 


PROBLEM. — Find  the  interest  due,  and   balance    this  ac- 
count : 


DR. 


F.  H.  Willis  in  acc't  with  E.  S.  Kennedy. 


CR 


1879. 

DoZ. 

Ct. 

1878. 

Dot. 

Ct. 

Jan. 

6 

To  check, 

170 

Dec. 

31 

By  balance  forward, 

325 

•"• 

13 

"  <      " 

480 

J879. 

16 

..  ^gf  (i 

96 

Jan. 

7 

"     cash  deposit, 

800 

" 

21 

"        " 

500 

'• 

20 

it                   44                         44 

175 

* 

50 

90 

240 

Interest  to  Jan.  31,  at  6  per  cent. 


DR. 


OPERATION. 


$170,  6^,  25 

da.=:$.708 

$325,  6« 

480,  " 

18 

"  =1.44 

8  0*0,  <; 

96,  " 

15 

a  __    2  4 

175,  " 

500,  " 

6 

"  =  .50 

240,  l( 

50,  « 

3 

"  ..=  .025 

$  1  5  4  0 

$1296 

$2.913 

$1296 

CR. 

31  da  .=$1.679 
24    "    =     3.20 
11    "    =        .32 


1 


$244 


=        .04 
$5.239 
$2.913 

$2.33~ 


.'.     Interest  due  =  $2.  33;    and  cash  balance  due  Willis  is 

$244  +  $2.33  =  $246.33,  Ans. 

SOLUTION. — The  interest  is  found  on  each  item  from  its  date  to 
the  day  of  settlement;  then  the  sum  of  the  items  and  the  sum 
of  the  interests  are  found  on  each  side.  The  difference  between 
the  sums  of  the  interests  is  equal  to  the  interest  due,  which, 
added  to  the  difference  between  the  sums  of  the  items;  gives  the 
balance  due. 

Hule. — 1.  Find  the  number  of  days  to  elapse  between  the 
date  when  each  item  is  due  and  the  date  of  settlement. 

2.  Find  the  sum  of  the  items  on  the  Dr.  side,  and  then  add 
to  each  item  its  interest,  if  the  item   is    due  before  the  date  of 
settlement,   or  subtract  it  if  due  after  the  date  of  settlement;  do 
the  same  with  Cr.  side. 

3.  The  difference  between  the  amounts   on   the   tivo   sides  of 
the  account  is  the  cash  balance. 


294 


RAY'S  HIGHER   ARITHMETIC. 


1.  Find  the  equated  time  and  cash  balance  of  the  follow- 
ing account,  July  7,  1876;  also  April  30,  1876,  interest 
6^  per  annum : 

Henry  Hammond. 
DR.  OK. 


1876. 

Jan. 

4 

To  Mdse,  at  3  mo. 

$900.10 

Feb. 

1 

"    4    " 

400.00 

ih 

700.50 

Mar. 

600.40 

April 
May 

8 
10 

'    Cash, 
'    Mdse,  at  30  da. 

500.20 
400.00 

June 

1 

4       "       "1  mo. 

100.60 

Equated  time,  May  15,  1876. 

Cash  balance  July  7,  1876,  $3633.62 

Cash  balance  April  30,  1876,  $3592.80 

REMARKS. — 1.  When,  in  forming  the  several  products,  the  cents 
are  50  or  less,  reject  them;  more  than  50,  increase  the  dollars  by  1. 

2.  The  cash  balance  is  the  sum  that  Henry  Hammond  will  be 
required  to  pay  in  settling  his  account  in  full  at  any  given  date. 

2.  Find  the  equated  time  and  cash  balance  of  the  follow- 
ing account,  Oct.  4,  1876,  money  being  worth  10^  per 
annum : 


DR. 


William  Smith. 


CR. 


1876. 

Dol. 

Ct. 

1876. 

Dot. 

Ct. 

Jan. 

1 

To  Mdse, 

800 

00 

Jan. 

Hi 

By  Cash, 

400 

00 

" 

16 

4       "       at  30  da. 

180 

30 

" 

28 

200 

00 

Feb. 

14 

"    60    " 

400 

60 

Feb. 

15 

Bills  Rec.  at  60  da. 

180 

30 

Mar. 

25 

*' 

500 

00 

28 

Cash, 

100 

00 

April 

1 

Cash, 

800 

00 

Mar. 

30 

Bills  Rec.  at  90  da. 

450 

00 

May 

7 

Mdse, 

600 

00 

April 

14 

Cash, 

400 

60 

*• 

•2\ 

at  60  da. 

700 

00 

May 

1 

*• 

500 

00 

June 

10 

200 

00 

15 

Bills  Rec.  at  1  mo. 

680 

00 

14 

15 

1        "    90  da. 

2000 

00 

June 

10 

Cash, 

300 

00 

July 

12 

500 

00 

July 

1!> 

70(3 

OJ 

Aug. 

4 

14    2  mo. 

1000 

10 

Aug. 

10 

Bills  Rec.  at  20  da. 

200 

00 

Sept. 

1 

Cash, 

150 

00 

Oct. 

3 

1100 

00 

Equated  time,  June  18,  1876;  cash  balance,  $2389.70 

REMARK. — All  written  obligations,  of  whatever  form,  for  which 
a  certain  amount  is  to  be  received,  are  called  Bills  Receivable. 


SETTLEMENT  OF  ACCOUNTS. 


295 


3.  Find    the    equated    time    and    the    cash    balance    of 
account,  March  31,  1876,  interest  10%  per  annum. 


DR. 


George  Cummings. 


CR. 


1876. 

Dot. 

Ct. 

1875. 

Dot. 

Ct. 

Jan. 

2 

To  Cash, 

300 

00 

Dec. 

1 

By  Mdse,  at  1  mo.  ' 

583 

00 

1876. 

M 

21 

it       44 

194 

00 

Feb. 

3 

•'        "       "     " 

40 

00 

Mar. 

4 

150 

00 

Mar. 

30 

130 

00 

Equated  time  Feb.   11,  1876;  cash  balance,  $110.48 

Find    the    interest    due,   and    balance    the    following   ac- 
counts : 

A.  L.  Morris  in  acc't  with  T.  J.  Fisher  &  Co. 
4.     DR.  CR. 


1871. 

Do!. 

Ct. 

1870. 

Dol. 

Ct. 

Jan. 

13 

To  check, 

350 

Dec. 

31 

By  bal.  from  old  acc't, 

813 

64 

" 

'22 

44                     44 

275 

1871. 

Feb. 

26 

(I                   4< 

100 

Feb. 

4 

"    cash, 

120 

May 

1 

'"                   " 

400 

Mar. 

17 

"        «* 

500 

June 

23 

108 

25 

May 

31 

84 

50 

Interest  to  June  30,  at  6  per  cent. 
Int.  due  Morris,  $13.34;  bal.  due  Morris,  S298.23 


5.    DR. 


Wm.   White  in  acc't   with  Beach  &  Berry. 


CR. 


1875. 

Dol. 

Ct. 

1875. 

Dol. 

Ct. 

July 

2 

lo  check, 

212 

50 

June 

30 

By  bal.  from  old  acc't, 

1102 

50 

" 

20 

66 

July 

6 

'     cash  deposit, 

50 

Aug. 

7 

235 

" 

15 

'        •«          " 

95 

25 

300 

Aug. 

9 

4                        44                              (4 

168 

75 

Sept. 

5 

110 

Sept. 

18 

4                    4.                         44 

32 

•« 

11 

46 

40 

Oct. 

3 

«                    41                         44 

79 

90 

27 

454 

25 

Interest  to  Oct.  12,  at  10  per  cent. 
Int.  due  White,  $19.68;  bal.  due  White,  $123.68 


296 


It  AY'S  HIGHER  ARITHMETIC. 


ACCOUNT  SALES. 

330.  1.  An  Account  Sales  is  a  written  statement  made 
by  an  agent  or  consignee   to  his  employer  or  consignor,  of 
the  quantity  and  price  of  goods  sold,  the  charges,  and  the 
net  proceeds. 

2.  Guaranty    is    a    charge    made    to    secure    the   owner 
against  loss  when  the  goods  are  sold  on  credit. 

3.  Storage  is  a  charge  made  for  keeping  goods,  and  is  usu- 
ally reckoned  by  the  week  or  month  on  each  piece  or  article. 

4.  In  Account  Sales  the  charges  for  freight,  commission, 
etc.,   are    the    Debits,   and    the   proceeds   of   sales    are    the 
Credits;    the    Net    Proceeds    is    the   difference    between    the 
sums  of  the  credits  and  debits. 

331.  Account  Sales  are  averaged  by  the  following  rule : 

Rule. — 1.  Average  the  sales  alone;  this  result  will  be  the 
date  to  be  given  to  the  commission  and  guaranty. 

2.  Make  the  sales  the  Or.  side,  and  the  charges  the  Dr. 
side,  and  find  the  equated  time  for  paying  tJie  net  proceeds. 

1.  Find  the  equated  time  of  paying  the  proceeds  on  the 
following  account  of  Charles  Maynard: 

Charles  Maynard?  s  Consignment. 


187G. 
Aug. 

191 

By  J.  Barnes 

,  at  10  da  

Dol. 
50 

Ct. 
60 

14 

Cash  .      .  ..         

800 

00 

i 

24 

(C                   <t 

Bills  Receivable,  at  30  da  

850 

00 

< 

*>9 

ii           ii 

Cash  

210 

00 

t 

<W 

4i           « 

George  Hand 

4900 

II!) 

i 

31 

((                   U 

Bills  Receivable,  at  20  da  

1400 

00 

1876. 
Aug. 

10 

To  Cash  paid 

CHARGES. 

Freight  .-....$  75.00 

8210 

GO 

SI 

Storage        .    .                                           10.00 

it 

31 

n       t(         it 

Insurance  Y&  per  cent  10.26 

« 

31 

14                 (4                     t( 

Commission  2X>  per  cent  205.25 

300 

^ 

Net  proceeds 

due  Maynard  

7910 

09 

Equated  time  Sept.  4,  1876. 


SETTLEMENT  OF  ACCOUNTS. 


297 


2.  Make  an  account  sales,  and  find  the  net  proceeds  and 
the  time  the  balance  is  due  in  the  following: 

William  Thomas  sold  on  account  of  B.  F.  Jonas  2000 
bu.  wheat  July  8,  1876,  for  $2112.50;  July  11,  300  bu. 
wheat,  and  took  a  20  day  note  for  $362.50  ;  paid  freight 
July  6,  1876,  $150.00;  July  11,  storage,  $6;  drayage,  $5; 
insurance,  $4;  commission,  at  2J%,  $61.87;  loss  and  gain 
for  his  net  gain,  $11.57 

Net  proceeds,  $2236.56;  due  July  12,  1876. 


3. 


Mdse.  Co. 


187(5. 
Jujy 

1876. 
July 

18 
24 
30 

15 
15 

30 
30 
30 
30 

By  Note    at  20  da  ... 

Dot. 

120 
60 
55 

Ct. 

00 
00 
(JO 

"    25  "  

"    Cash     ..  ..                          

CHARGES. 

To  Cash  paid     Freight       

$  33  00 

'        "         "        Drayage  ,  

4.00 

4        "          "        Insurance  

1.50 

"        "         "       Storage  

'       "         "        Commission,  at  2H  per  cent  . 

3.00 
529 

C.  V.  Carries  's  net  proceeds  

62.73 

Less  our  /•£  net  loss 

$109.52 
3  93 

§105.59 

Find  C.  V.  Carnes's  net   proceeds  if  paid  Jan.   1,  1877, 

money  being  worth  10^  per  annum ;  also  the  equated  time. 

Equated  time  Aug.  19,  1876;  net  proceeds,  $65.03 


STORAGE  ACCOUNTS. 

332.  Storage  Accounts  are  similar  to  bank  accounts, 
one  side  showing  how  many  barrels,  packages,  etc.,  are 
received,  and  at  what  times ;  the  other  side  showing  how 
many  have  been  delivered,  and  at  what  times. 

Storage  is  generally  charged  at  so  much  per  month  of 
30  days  on  each  barrel,  package,  etc. 


298 


RAY'S  HIGHER  ARITHMETIC. 


1.    Storage  to  Jan.  81,  at  5  ct.  a  bbl.  per  month. 

RECEIVED.  DELIVERED. 


3876. 

Bbl. 

Balance  on  hand. 

Days. 

Products. 

1876. 

Bbl. 

January 

2 

200 

January 

10 

110 

5 

150 

13 

90 

7 

30 

17 

20 

10 

120 

20 

115 

14 

80 

25 

Ji° 

17 

150 

27 

20 

75 

30 

100 

24 

60 

31 

28 

200 

Storage,  $19.25;  bbl.  on  hand,  418. 
2.  Storage  to  Feb.  20,  at  5  ct  a  bbl.  per  month. 


RECEIVED. 


DELIVERED. 


Bbl. 

Balance  on  hand. 

Days. 

Products. 

Bbl. 

January 
February 

31 
4 

418 
250 

February 

5 
10 

100 

80 

9 

120 

12 

220 

12 

100 

14 

140 

16 

30 

18 

90 

20 

288 

Storage,  $15.85;  bbl.  on  hand,   0. 


VIII.    COMPOUND   INTEREST. 
DEFINITIONS. 

333.  1.  Compound  Interest  is  interest  computed  both 
upon  the  principal  and  upon  each  accrued  interest  as 
additional  principal. 

2.  Annual  Interest  is  the  gain  of  a  principal  whose  yearly 
interests  have  become  debts  at  simple  interest ;  but,  distinct 
from  this,  Compound  Interest  is  the  whole  gain  of  a  principal, 
increased  at  the  end  of  each  interval   by  all  the  interest  draw^i 
during  that  interval. 

3.  The  final  amount  in  Compound  Interest  is  called  tho 
compound  amount. 


COMPOUND  INTEREST.  299 

EXAMPLE.— Let  the  principal  be  $1000,  the  rate  per  cent  6.  The 
first  year's  interest  is  $60.  If  this  be  added  as  a  new  debt,  the  prin- 
cipal will  become  $1060,  and  the  second  year's  interest  $63.60;  in 
like  manner,  the  third  principal  is  $1123.60,  and  a  third  year's 
interest  $67.416;  then,  the  whole  amount  is  $1191.016,  and  the 
whole  gain,  $191.016 

REMARK. — If  the  above  debt  be,  at  the  same  rate,  on  annual  interest 
(Art.  304),  the  whole  amount  will  be  $1190.80,  and  the  whole  gain 
$190.80;  the  difference  is  the  interest  of  $3.60  (an  interest  upon  an 
interest  debt)  for  one  year,  $.216 

Compound  Interest  has  four  cases. 


CASE    I. 

334.  Given  the  principal,  rate,  and  time,  to  find  the 
compound  interest  and  amount. 

PROBLEM. — Find  the  compound  amount  of  $1000,  in  4 
years,  at  2^  per  annum 

SOLUTION. — Multiplying  the  principal,  $1000,  by  1.02,  the  num- 
ber expressing  the  amount  of  $1  for  a  year,  we  have  the  first  year's 
amount,  $1020.  Continuing  the  use  of  the  factor  1.02  until  the 
fourth  product  is  obtained,  we  have  for  the  required  amount, 
$1082.43216  The  same  numerical  result  would  have  been  obtained 
by  taking  1.02  four  times  as  a  factor,  and  multiplying  the  product 
by  1000. 

REMARK. — Compound  Interest  may  be  payable  semi-annually  or 
quarterly,  and  in  such  cases  the  computation  is  made  by  a  multi- 
plication similar  to  the  last. 

EXAMPLE. — Let  the  debt  be  $1000,  and  let  the  interest  be  com- 
pounded at  2^>  quarterly.  In -one  year  there  are  four  intervals, 
and,  as  seen  in  the  last  process,  the  year's  amount  is  $1082.43216 
The  real  gain  on  each  dollar  is  $.0824+,  or,  a  fraction  over  826-^. 
According  to  the  usual  form  of  statement,  this  debt 'is  compounding 


300  RA  Y'S  HIGHER  ARITHMETIC, 

"  at  8<fi  per  annum,  payable  quarterly."  But  this  must  not  be  under- 
stood as  an  exact  statement  of  the  real  gain  ;  for,  when  the  quarterly 
rate  is  2^,,  the  annual  rate  exceeds  8^,  and  when  the  annual  rate  is 
exactly  8<fa,  the  quarterly  rate  is  1.943^,  nearly. 

REMARK. — To  ascertain  the  true  rate  for  a  smaller  interval  when 
the  yearly  rate  is  given,  requires  to  separate  the  yearly  multiplier 
into  equal  factors.  Thus  the  true  half-year  rate,  when  the  annual 
rate  is  21^,  is  found  by  separating  1.21  into  two  equal  factors,  1.10 
X  1.10;  and  the  quarterly  rate,  when  the  annual  is  8^,  is  found  by 
separating  1.08  into  four  factors,  each  nearly  1.01943  It  is  true  that 
this  process  is  rarely  demanded,  and  that  it  is  very  tedious  when 
'the  intervals  are  small;  but,  in  a  proper  place,  it  will  be  a  useful 
exercise.  (See  Art.  389). 

Rule. — Find  the  amount  of  the  principal  for  the  first  in- 
terval, at  the  rate  for  that  interval,  and  in  like  manner  treat 
the  whole  debt,  at  the  end  of  each  interval,  as  a  principal  at 
simple  interest  through  the  following  interval  or  part  of  an 
interval;  the  result  will  be  the  compound  amount.  To  find  the 
compound  interest,  deduct  the  original  debt  from  the  compound 
amount. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  compound  amount  and  interest  of  83850,  for 
4  yr.  7  mo.  16  da.,  at  5%,  payable  annually. 

$4826.59,  and  $976.59 

2.  The  compound  interest  of  $13062.50,  for  1  yr.  10  mo. 
12  da.,  at  8%,  payable  quarterly.  $2082.25 

3.  The  compound   amount  of  $1000,  for  3   yr.,  at  10%, 
payable  semi-annually.  $1340.10 

4.  What  sum,  at  simple  interest,  6^,  for  2  yr.  5  mo.  27 
da.,  amounts  to  the   same  as  $2000,  at  compound  interest, 
for  the  same  time  and  rate,  payable  semi-annually? 

$2016.03 

5.  What   is   the    difference    between  the  annual   and  the 
compound  interest  of  $5000  in  6  years,  6%  per  annum? 

$22.596 


COMPOUND  INTEREST.  301 

6.  Required  the  amount  of  $1000,  at  compound  interest, 
21%,  for  2yr.  6  mo.  $1617.83 

REMARKS. — 1.  In  obtaining  the  answer  to  the  last  problem,  the 
compound  amount  at  the  end  of  the  second  full  interval  is  treated 
as  a  sum  at  simple  interest  for  6  months.  But,  calculated  at  a  true 
half-year  rate,  the  amount  is  only  $1610.51,  and  this  sum  continued 
at  the  same  true  rate  for  the  remaining  half  year  will  amount  to  the 
same  sum  which  the  debt  would  have  reached  in  a  full  interval ; 
for, 

$1464.10  X  1-21  =  $1771.561  ;  and 

$1464.10,  for  2  intervals,  at  10%  comp.  int.  = 

$1464.10  X  1.10  X  1.10  =  $1771.561 

2.  Let  the  student  carefully  note  that  the  interest  drawn  during 
a   year   may   be    considered    as    the   sum   of    interests    compounded 
through  smaller  intervals,  at  smaller  rates.     Thus,  6^  a  year  may 
be  regarded  as  the  sum  of  interest  compounded   through  quarterly 
intervals   at  1.46.7^,  through  monthly  intervals  at  .487^,,  or  daily 
at   .016$),  approximately.      Suppose  7  months  have   passed   since 
interest  began.     The  year  may  be  taken  as  a  period  of  12  intervals, 
and  the  interest  as  having  been  compounded  through  7  of  them,  at 
.487  ^j.      If    the   amount  then  reached  be  continued  at   compound 
interest   through   the   other   5   intervals,  the    amount   will    be   the 
same  as  that  of  the  debt   continued  to  the  end  of  the  year  at  the 
full  rate. 

3.  In  this  view,  the  statement  may  be  made,  general,  that  the  worth 
of  the  debt  at  any  point  in  a  year,  is  the  principal,  which,  compounding  at 
a  true  partial  rate  for  the  remaining  fraction  of  a  year,  will  amount  to  the 
same  sum  as  the  debt  continued  through  that  year  at  the  annual  rate. 

This  is  in  strict  accordance  with  the  results  obtained  by  Algebra, 
but  in  the  common  calculations  of  Arithmetic,  the  interest  is  added 
to  the  debt,  and  the  rate  divided,  only  according  to  the  statements 
'•payable  annually,"  "payable  quarterly,"  etc. 


CALCULATION  BY  TABLES. 

335.  When  the  intervals  are  many,  the  actual  multipli- 
cations become  laborious;  and,  therefore,  tables  of  compound 
interest  have  been  prepared  to  shorten  the  work. 


302 


RAY'S  HIGHER  ARITHMETIC. 


Amount  of  $1  at  Compound  Interest  in  any  number  of  years,  not 
exceeding  fifty-five. 


Yrs. 

2  per  cent. 

2%  per  cent. 

3  per  cent. 

1%  per  cent. 

4  per  cent. 

4>2  per  cent. 

2 
3 

5 

1.0200  0000 
1.0404  0000 
J.0612  0800 
1.0824  3216 
1.1040  8080 

1.0250  0000 
1.0506  2oOO 
1.0768  9062 
1.1038  1289 
1.1314  0821 

1.0300  0000 
1.0609  (XXX) 
1.0927  2700 
1.1255  0881 
1.1592  7407 

1.0350  0000 
1.0712  2500 
1.1087  1787 
1.1475  2300 
1.1876  8631 

1.0400  0000 
1.0816  0000 
1.1248  6400 
1.1698  5856 
1.2166  5290 

1.04.50  0000 
1.0920  2500 
1.1411  6612 
1.1925  1S60 
1.2461  8194 

6 

7 
8 
9 
10 

1.1261  6242 
1.1486  8567 
1.1716  5938 
1.1950  9257 
1.2189  9442 

1.1596  9342 
1.1886  8575 
1.2184  0290 
1.2488  6297 
1.2800  8454 

1.1940  5230 
1.2298  7387 
1.2667  7008 
1.3047  7318 
1.3439  1638 

1.2293  5533 
1  .2722  7926 
1.3168  0904 
1.3628  9735 
1.4105  9876 

1.2653  1902 
1.3159  3178 
1.3685  6905 
1.4233  1181 
1.4802  4428 

1.3022  6012 
1.360S  61  Si 
1.4221  0061 
1.4860  9514 
1.5529  6942 

11 
12 
13 
14 
15 

1.2433  7431 
1.2682  4179 
1.2936  0663 
1.3194  7876 
1.3458  6834 

1.3120  8666 
1.3448  8882 
1.3785  1104 
1.4129  7382 
1.4482  9817 

1.3842  3387 
1.4257  6089 
1.4685  3371 
1.5125  8972 
1.5579  6742 

1.4599  6972 
1.5110  6866 
1.5639  5606 
1.6186  9452 
1.6753  4883 

1.5394  5406 
1.6010  3222 
1.6650  7351 
1.7316  7645 
1.8009  4351 

1.6228  5305 
1.6958  8143 
1.7721  9610 
1.85  19  4492 
1.9352  8244 

16 
17 

18 
19 
20 

1.3727  8570 
1.4002  4142 
1.4282  4625 
1.4568  1117 
1.4859  4740 

1.4845  0562 
1.5216  1826 
1.5596  5872 
1.5986  5019 
1.6386  1644 

1.6047  0644 
1.6528  4763 
1.7024  3306 
1.7535  0605 
1.8061  1123 

1.7339  8604 
1.79J6  7555 
1.8574  8920 
1.9225  0132 
1.9897  8886 

1.8729  8125 
1.9479  <X)50 
2.0258  1652 
2.1068  4918 
2.1911  2314 

2.0223  7015 
2.1133  7681 
2.2084  7877 
2.3078  6031 
2.4117  1402 

21 
22 
23 
24 
25 

1.5156  6634 
1.5459  7967 
1.5768  9926 
1.6084  3725 
1.6406  0599 

1.6795  8185 
1.7215  7140 
1.7646  1068 
1.8087  2595 
1.8539  4410 

1.8602  9457 
1.9161  0341 
1.9735  8651 
2.0327  9411 
2.0937  7793 

2.0594  3147 
2.1315  1158 
2.2061  1448 
2.2833  2849 
2.3632  4498 

2.2787  6807 
2.3699  1879 
2.4647  1555 
2.5633  0417 
2.6658  3633 

2.5202  4116 
2.6336  5201 
2.7521  6635 
2.87(50  1383 
3.(X)54  3446 

26 

27 
28 
29 
30 

1.6734  1811 
1.7068  8H48 
1.7410  2421 
1.775S  4469 
J.S11!  6158 

1  .9002  9270 
1.9478  0002 
1.9964  9502 
2.0464  0739 
2.0975  6758 

2.1565  9127 
2.2?12  8901 
2.28,9  2768 
2.3565  6551 
2.4272  6247 

2.4459  5856 
2.5315  6711 
2.6201  7196 
2.7118  7798 
2.8067  9370 

2.7724  6979 
2.8833  6858 
2.9987  0332 
3.1186  5145 
3.2433  9751 

3.1406  7901 
3.2820  0956 
3.4296  9999 
3.5840  3649 
3.7453  1813 

31 
32 
33 
34 
35 

1.8475  8882 
1.8845  4059 
1.9222  3140 
1.9606  7603 
1,9998  8955 

2.1500  0677 
2.2037  5694 
2.2588  5086 
2.3153  2213 
2.3732  0519 

2.5000  8035 
2.5750  8276 
2.6523  3524 
2.7319  0530 
2.8138  '6245 

2.9050  3148 
3.0067  0759 
3.1119  4235 
3.2208  6033 
3.3335  9045 

3.3731  3341 
3.5080  5875 
3.6483  8110 
3.7943  1634 
3.9460  8899 

3.9138  5745 
4.0899  8104 
4.2740  3018 
4.4663  6154 
4.6673  47S1 

36 
37 
38 
39 
40 

2.0398  8734 
2.0S06  8509 
2.1222  9879 
2.1647  4477 
2.2080  3966 

2.4325  3532 
2.4933  4870 
2.5556  8242 
2.6195  74-18 
2.6850  6384 

2.8982  7833 
2.9852  2668 
3.0747  8348 
3.1670  2698 
3.2620  3779 

3.4502  6611 
3.5710  2-343 
3.6960  1132 
3.8253  7171 
3.9592  5972 

4.1039  3255 

4.2680  8986 
4.4388  1345 
4.6163  6599 
4.8010  2063 

4.8773  7846 
5.0968  6049 
5.3262  1921 
5.5658  9908 
5.8163  6454 

41 
42 
43 
44 
45 

2.2522  0046 
2.2972  4447 
2.3431  8936 
2.3900  5314 
2.4378  5421 

2.7521  9043 
2.8209  9520 
2.8915  2008 
2.9638  0808 
3.0379  0328 

3.3598  9893 
3.4606  9589 
3.5645  1677 
3.6714  5227 
3.7815  9584 

4.0978  3381 
4.2412  5799 
4.3897  0202 
4.5433  41  M) 
4.7023  5855 

4.9930  6145 
5.1927  8391 
5.4004  9527 
5.616-5  1508 
5.8411  7568 

6.0781  0094 
6.3516  1548 
6.6374  3S18 
6.9361  2290 
7.2482  4843 

46 
47 
48 
49 
50 

2.4866  1129 
2.5363  4351 
2.5870  7039 
2.6388  1179 
2.6915  8803 

3.1138  5086 
3.1916  9713 
3.2714  8956 
3.3532  7680 
3.4371  OS72 

3.8950  4372 
4.0118  9503 
4.1322  5188 
4.2562  1944 
4.3839  0602 

4.8669  4110 
5.0372  8404 
5.2135  8898 
5.3960  6459 
5.5849  26S6 

6.0748  2271 
6.3178  1562 
6.5705  2824 
6.8333  41)37 
7.1066  8335 

7.5744  1961 
7.9152  6849 
8.2714  5557 
8.6436  7107 
9.0326  3627 

51 
52 
53 
54 
55 

2.7454  1979 
2.8003  2«19 
2.8563  3475 
2.9134  6144 
2.9717  3067 

3.5230  3644 
3.6111  1235 
3.7013  9016 
3.7939  2491 
3.8887.7303 

4.5154  2320 
4.6508  8590 
4.7904  1247 
4.9341  2485 
5.0821  4859 

5.7803  9930 
5.9S27  1327 
6.1921  0824 
6.4088  3202 
6.6331  4114 

7.3909  5068 
7.6865  8S71 
7.9940  5226 
8.3138  1435 
8.6463  6692 

9.4391  0490 
9.8638  6463 
10.3077  3«5:j 
10.7715  8677 
11.2563  0817 

Subtract  $1  from  the  Amount  in  this  Table  to  fnd  the  Interest. 


COMPOUND  INTEREST. 


303 


Amount  of  $1  at  Compound  Interest  in  any  number  of  years,  not 
exceeding  f/ty-five. 


Yrs. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

9  per  cent. 

10  per  cent. 

1 
2 
3 
4 
5 

1.0500  000 
1.1025  000 
1.1576  250 
1.2155  003 
1.2762  816 

1.0600  000 
1.1236  000 
1.1910  160 
1.2624  770 
1.3382  256 

1.0700  000 
1.1449  000 
1.2250  430 
1.3107  960 
1.4025  517 

1.0800  OIK) 
1.1064  0(H) 
1.2597  120 
1.3004  890 
1.4693  281 

1  .0900  000 
1.1881  000 
1.2950  290 
1.4115  810 
1.5386  240 

1.1000  000 
J  .2100  000 
1.3310  OliO 
1.4011  0(,0 
1.61U5  1(JO 

6 
7 
8 
9 
10 

1.3400  956 
1.4071  004 
1.4774  554 
1.5513  282 
1.6288  946 

1.4185  191 

1.5036  303 
1.5938  481 
1.6894  790 
1.7908  477 

1.5007  304 
1.6057  815 
1.7181  862 
1.8384  592 
1.9671  514 

1.5868  743 
1.7138  243 
1.8509  302 
1.995)0  040 
2.1589  250 

1.0771  001 
1.8280  391 
1.9925  020 
2.1718  933 
2.3073  637 

1.7715  610 
1.9487  171 
2.1435  888 
2.3579  477 
2.5937  425 

11 
12 
13 
14 
15 

1.7103  394 
1.7958  563 
1.8856  491 
1.979!)  316 
2.0789  282 

1.8982  986 
2.0121  965 
2.1329  283 
2.2609  040 
2.3965  582 

2.1048  520 
2.2521  916 

2.4098  450 
2.5785  342 
2.7590  315 

2.3316  390 
2.5181  701 
2.7196  237 
2.9371  936 
3.1721  691 

2.5804  264 
2.8126  648 
3.0058  046 
3.3417  270 
3.6424  825 

2.8531  167 
3.1384  284 
3.4522  712 
3.7974  983 
4.1772  482 

16 
17 

18 
19 

20 

2.1828  746 
2.2920  183 
2.4060  J92 
2.5269  502 
2.6532  977 

2.5403  517 

2.6927  728 
2.8543  392 
3.0255  995 
3.2071  355 

2.9521  638 
3.1588  152 
3.3799  323 
3.6165  275 
3.86U6  845 

3.4259  426 
3.7000  181 
3.9960  195 
4.3157  Oil 
4.6609  571 

3.9703  059 
4.3276  334 
4.7171  204 
5.1416  613 
5.6044  108 

4.5949  730 
5.0544  703 
5.5599  173 
6.1159  090 
6.7275  000 

21 
22 
23 
24 
25 

2.7859  626 
2.9252  607 
3.0715  238 
3.2250  993 
3.3863  549 

3.3995  636 
3.6035  374 
3.8197  497 
4.0489  346 
4.2918  707 

4.1405  624 
4.4304  017 
4  7405  299 
5.0723  670 
5.4274  326 

5.0338  337 
5.4365  404 
5.8714  637 
6.3411  807 
6.8484  752 

6.1088  077 
6.0580  004 
7.2578  745 
7.91  10  832 
8.6230  807 

7.4002  499 
8.1402  749 
8.9543  024 
9.8497  327 
10.8347  059 

26 
27 

28 
29 
30 

3.5556  727 
3.7334  563 
3.9201  291 
4.1161  356 
4.3219  424 

4.5493  830 
4.8233  459 
5.1116  867 
5.4183  879 
5.7434  912 

5.8073  529 
6.2138  676 
6.6488  384 
7.J142  571 
7.6122  550 

7.3963  532 
7.9880  615 
8.627  L  064 
9.3172  749 
10.0626  569 

9.3991  579 
10.2450  821 
11.1071  395 
12.1721  821 
13.2676  785 

11.9181  765 
13.1099  942 
14.4209  936 
15.8030  930 
17.4494  023 

31 
32 
33 
34 
35 

4.5380  395 
4.7649  415 
5.0031  885 
5.2533  480 
5.5160  154 

6.0881  006 
6.45:53  867 
6.8405  899 
7.2510  253 

7.6860  868 

8.1451  129 
8.7152  708 
9.3253  398 
9.9781  135 
10.6765  815 

10.8676  694 
11.7370  830 
12.6760  496 
13.6901  336 
14.7853  443 

14.4617  695 
15.7033  288 
17.1820  284 
18.7284  109 
20.4139  679 

19.1943  425 
21.1137  708 
23.2251  544 
25.5476  099 
28.1024  309 

36 
37 

38 
39 
40 

5.7918  16.1 
6.0814  069 
6.38">4  773 
6  7047  512 
7.0399  887 

8.1472  520 
8.6360  871 
9.1542  524 
9.7035  075 
10.2857  179 

11.4239  422 
12.2236  181 
13.0792  714 
13.9948  204 
14.9744  578 

15.9681  718 
17.2456  256 
18.6252  756 
20.1152  97? 
21.7245  215 

22.2512  250 
24.2538  353 
20.4300  805 
28.8159  b!7 
31.4094  200 

30.9120  805 
34.0039  480 
37.4043  434 
41.1447  778 
45.2592  556 

41 
42 
4:j 
44 
45 

7.3919  882 
7.7615  876 
8.1496  669 
8.5571  503 
8.9850  078 

10.9028  610 
11.5570  327 
12.2504  546 
12.9854  819 
13.7646  108 

16.0226  699 
17.1442  568 
18.3443  548 
19.0284  596 
21.0024  518 

23.4024  832 
25.3394  819 
27.3660  404 
29.5559  717 
31.9204  494 

34.2362  679 
37.3175  320 
40.6701  098 
44.33(59  597 
48.3272  861 

49.7851  811 
54.7636  992 
00.2400  692 
00.2040  701 
72.8904  837 

40 
47 

48 
49 
50 

9.4342  582 
9.9059  711 
10.4012  697 
10.9213  331 
11.4673  998 

14.5901  875 
15.4659  167 
16.3938  717 
17.3775  040 
18.4201  543 

22.4726  234 
24.0457  070 
25.72S9  065 
27.5299  300 
29.4570  251 

34.4740  853 
37.2320  122 
40.2105  731 
43.4274  190 
40.9016  125 

52.6707  419 
57.4176  486 
62.5852  370 
08.2179  083 
74.3575  201 

80.1795  321 
88.1974  853 
97.0172  338 
106.7189  572 
117.3908  529 

51 
52 
53 
54 
55 

12.0407  698 
12.6428  083 
13.2749  487 
13.9.583  961 
14.6356  309 

19.5253  635 
20.6968  853 
21.9386  985 
23.2550  204 
24.6503  216 

31.5190  168 
33.7253  480 
30.0801  224 
38.6121  509 
41.3150  015 

50.6537  415 

54.7000  408 
59.0825  241 
63.80!)!  200 
68.9138  561 

81.0496  969 
88.3441  090 
90.2951  449 
104.9617  079 
114.4082  616 

129.1299  382 
142.0429  320 
150.2472  252 
171.8719  477 
189.0591  425 

Subtract  $1  from  the  Amount  in  this  Table  to  find  the  Interest. 


304  RAY'S  HIGHER  ARITHMETIC. 

How  to  use  the  table  in  finding  the  Compound  Amount: 

1.  Observe  at  what  intervals  interest  is  payable,  and  also  the 
rate  per  interval. 

2.  If  the  number  of  full  intervals  can  be  found  in  the  year 
column,  note  the  sum  corresponding  to  it  in  the  column  under 
the  proper  rate;    multiply  this  sum,   or  its  amount  for  any  re- 
maining fraction  of  an  interval,  by  the  principal. 

3.  If  the    number    of  intervals   be    not  found   in   the   table, 
separate  the  whole  time  into  periods  which  are   each  within  the 
limits  of  the  table;  find    the    amount  of  the  principal  for  one 
of  them,  make  that  a  principal  for  the  next,  and  so  on,  till  the 
whole  time  has  been  taken  into  the  calculation. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  compound  amount  of  $750  for  17  yr.,  at  6%, 
payable  annually.  $2019.58 

2.  Of  $5428  for  33  yr.,  5^  annually.  $27157.31 

3.  The   compound   interest  of  $1800  for  14  yr.,  at  8%, 
payable  semi-annually.  $3597.67 

4.  If  $1000  is  deposited  for  a  child,  at  birth,  and  draws 
1%   compound   interest,  payable  semi-annually,  till   it  is  of 
age  (21  yr.),  what  will  be  the  amount?  $4241.26 

5.  Find  the  amount  of  $9401.50,  at  compound  interest  for 
19  yr.  4  mo.,  9%,  payable  semi-annually.  $51576.68 

6.  Find  the  compound  amount  of  $1000  for  100  yr.,  at 
10%,  payable  annually.  $13780612.34 

7.  The   compound   interest   of  $3600  for  15  yr.,  at  S%, 
payable  quarterly.  $8211.71 

8.  The   compound  interest  of  $4000  for  40   yr.,  at  5%, 
payable  semi-annually.  $24838.27 

9.  The  compound  interest  of  $1200  for  27  yr.  11  mo.  4  da., 
at  12%,  payable  quarterly.  $31404.74 


COMPOUND  INTEREST.  305 


CASE     II. 

336.  Given  the  principal,  rate,  and  compound  in- 
terest or  amount,  to  find  the  time. 

PROBLEM. — Find  the  time  in  which  $750  will  amount  to 
$2000,  the  interest  being  8  %,  payable  semi-annually. 

SOLUTION. — Since  a  compound  amount  is  found  by  multiplying 
the  principal  by  the  amount  of  $1,  we  here  reverse  that  process,  and 
say:  $2000  -f-  750  =  $2.66666666,  the  amount  of  $1,  at  4^>.  The 
number  next  lower,  in  the  4^  column,  is  $2.66583633,  the  amount 
for  25  intervals,  and  is  less  than  $2.66666666  by  $.00083033 

Since  the  amount  for  25  intervals  will,  according  to  the  table, 
gain  $.10663343  in  1  interval,  it  will  gain  $.00083303  in  such  a  frac- 
tion of  an  interval  as  the  latter  sum  is  of  the  former;  . 00083033 -5- 
.10663346  =  T|-g-,  nearly  ;  hence,  the  required  period  is  25  j\j  inter- 
vals of  6  mo.,  or,  12 -yr.  6  mo.  1  da.,  Ans. 

Rule. — 1.  Divide  the  amount  by  the  principal. 

2.  If  the  quotient  be  found  in  the  table  under  the  given  rate, 
the  years  opposite  will  be  the  required  number  of  intervals;   but 
if  not  found  exactly,   in  the   table,  take    the    number  next  less, 
noting  its  deficiency,  its  number  of  years,  and  its  gain  during 
a  full  interval. 

3.  Divide  the  deficiency  by  the  interval   gain,  and  annex  the 
quotient  to   the  number  of  full  intervals ;    the  result  will  be  the 
required  time. 


EXAMPLES  FOR  PRACTICE. 

In  what  time,  at  compound  interest,  will: 

1.  $8000  amount  to  $12000,  at  6%? 

6  yr.  11  mo,  15  da. 

2.  $5200  amount  to  $6508,  Q%,  payable  semi-annually? 

3  yr.  9  mo.  16  da. 

H.  A.  26. 


306  RAY'S  HIGHER  ARITHMETIC. 

3.  &12500  gain  $5500,  10%,  payable  quarterly? 

3  yr.  8  mo.  9  da. 

4.  $1  gain  91,  at  6,  8,  10  ft? 

11  yr.  10  mo.  21  da.;  9  yr.  2  da.;  7  yr.  3  mo.  5  da. 

5.  $9862.50  amount   to   $22576.15,  12$",  payable   semi- 
annual ly?  7  yr.  1  mo.  7  da. 

CASE   III. 

337.  Given  the  principal,  the  compound  interest  or 
amount,  and  the  time,  to  find  the  rate. 

PROBLEM.—  At  what  rate  will  $1000  amount  to  $2411.714 
in  20  years? 

SOLUTION.—  Dividing  $2411.714  by  1000,  we  have  $2.411714, 
which,  in  the  table,  corresponds  to  the  amount  of  $1  for  the  time, 
at 


Rule.  —  Divide  the  amount  by  the  principal;  search  in  the 
table,  opposite  the  given  number  of  full  intervak,  for  the  exact 
quotient  or  the  number  nearest  in  value;  if  the  time  contain 
also  a  part  of  an  interval,  find  the  amount  of  the  tabular  sum 
for  that  time,  before  comparing  with  the  quotient;  the  rate  per 
Gent  at  the  head  of  the  column  ivill  be  the  exact,  or  the  approx- 
imate rate. 


EXAMPLES  FOR  PRACTICE. 

At  what  rate,  by  compound  interest, 

1.  Will  $1000  amount  to  $1593.85  in  8  yr.?  .  6%. 

2.  $3600  amount  to  $9932.51  in  15  yr.?  7%. 

3.  $13200  amount  to  48049.58,  in  26  yr.  5  mo.  21  da.  ? 

5%. 

4.  $2813.50  amount  to  $13276.03,  in  17  yr.  7  mo.  14  da., 
interest  payable  semi-annually  ?  9%, 


COMPOUND  INTEREST.  307 

5.  $7652.18  gain  $17198.67,  interest  payable  quarterly, 
in  11  yr.  11  mo.  3  da.?  10%. 

6.  Any  sum  double  itself  in  10,  15,  20  yr.  ? 

1st,  between  1%  and  8^;  2d,  nearly  5^; 
3d,  little  over  3|%. 

CASE    IV. 

338.     Given  the  compound    interest  or  amount,  the 
time,  and  the  rate,  to  find  the  principal. 

PROBLEM.—  What    principal   will    yield    $31086.78    com- 
pound interest  in  40  yr.,  at 


SOLUTION.—  In  40  yr.  $1  will  gain  $20.7245215,  at  the  given  rate  ; 
the  required  principal  must  contain  as  many  dollars  as  this  interest 
is  contained  times  in  the  given  interest;  $31086.78-^-20.7245215  = 
$1500,  Am. 

Rule.  —  Divide  the  given  interest  or  amount  by  the  interest 
or  amount  of  $1  for  the  given  time  and  at  the  given  rate;  the 
quotient  will  be  the  required  principal. 

REMARK.  —  If  the  amount  be  due  at  some  future  time,  the  prin- 
cipal is  the  present  worth  at  compound  interest,  and  the  difference 
between  the  amount  and  present  worth  is  the  compound  discount. 


EXAMPLES  FOR  PRACTICE. 

What  principal,  at  compound  interest, 

1.  Will  yield  $52669.93  in  25  yr.,  6^?  $16000. 

2.  Will  gain  $1625.75  in  6  yr.  2  mo.,  1%,  payable  semi- 
annually?  $3075. 

3.  Will  yield  $3598.61  in  3  yr.  6  mo.  9  da.,  W%,  payable 
quarterly?  $8640. 

4.  Will  yield  $31005.76  in  9£  yr.,  at  8%,  payable  semi- 
annually?  $28012.63 


308  RAY'S  HIGHER  ARITHMETIC. 

5.  Will  amount  to  $27062.85  in  7  yr.,  at  4%  ? 

$20565.54 

6.  What  is  the  present  worth  of  $14625.70,  due  in  5  yr. 
9  mo.,  at  6^  compound  interest,  payable  semi-annually  ? 

$10409.77 

7.  What  is  the  compound  discount  on  $8767.78,  due  in 
12  yr.  8  mo.  25  da.,  5^?  $4058.87 


IX.    ANNUITIES* 

DEFINITIONS. 

339.     1.    An    Annuity  is  a  sum  of  money  payable  at 
yearly  or  other  regular  intervals. 

(  1.  Perpetual  or  Limited. 
Annuities  are — <  2.  Certain  or  Contingent. 

(.3.  Immediate  or  Deferred. 

2.  A  Perpetual  Annuity,  or  a  Perpetuity,  is  one  that 
continues  forever. 

3.  A  Limited  Annuity  ceases  at  a  certain  time. 

4.  A  Certain  Annuity  begins  and  ends  at  fixed  times. 

5.  A  Contingent  Annuity  begins  or  ends  with  the  hap- 
pening of  a  contingent  event — as  the  birth  or  the  death  of  a 
person. 

6.  An  Immediate  Annuity  is  one  that  begins  at  once. 

7.  A  Deferred  Annuity,  or  an  Annuity  in  Reversion, 
is  one  that  does  not  begin  immediately ;  the  term  of  the  re- 
version may  be  definite  or  contingent. 


*  Si  nee  the  problems  in  annuities  may  be  classed  under  the 
Applications  of  Percentage,  the  subject  is  presented  here,  instead 
of  being  placed  after  Progression ;  however,  those  who  prefer  may 
omit  this  chapter  until  after  Progression  has  been  studied. 


ANNUITIES.  309 

8.  An   annuity   is   Forborne  or  in  Arrears  if  not  paid 
when  due. 

9.  The  Forborne  or  Final  Value  of  an  annuity  is  the 
amount  of  the  whole  accumulated  debt  and  interest,  at  the 
time  the  annuity  ceases. 

10.  The  Present  Value  of  an  annuity  is  that  sum,  which, 
put  at  interest  for  the  given   time   and  at  the  given   rate, 
will  amount  to  the  final  value. 

11.  The  value  of  a  deferred  annuity  at  the  time  it  com- 
mences, may  be  called  its  Initial  Value ;  its  Present  Value 
is  the  present  worth  of  its  initial  value,  at  an  assumed  rate 
of  interest. 

12.  The  rules  for  annuities  are  of  great  importance;  their 
practical  applications  include  leases,  life-estates,  rents,  dowers, 
pensions,  reversions,  salaries,  life  insurance,  etc. 

REMARK. — An  annuity  begins,  not  at  the  time  the  first  payment 
is  made,  but  one  interval  before;  if  an  annuity  begin  now,  its  first 
payment  will  be  a  year,  half-year,  or  quarter  of  a  year  hence,  accord- 
ing to  the  interval  named. 


CASE  I. 

340.     Given    the    payment,   the    interval,4  and    the 
rate,  to  find  the  initial  value  of  a  perpetuity. 

PROBLEM. — What  is  the  initial  value  of  a  perpetual  lease 
of  $250  a  year,  allowing  6^  interest  ? 

OPERATION. 

SOLUTION. — The  initial  value  must  $1 

be  the  principal,  which,  at  6^,  yields          .  0  6 
$250  interest  every  year ;  it  is  found,          .06)250.0000 
by  Art.  300,  $4166.66 f , Am. 

Rule. — Divide  the  given  payment  by  the  interest  of  $1  for 
one  interval  at  the  proposed  rate. 


310  RA  Y'S  HIGHER  ARITHMETIC. 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  initial  value  of  a  perpetual  leasehold  of 
$300  a  year,  allowing  6^  interest?  $5000. 

2.  What  must  I  pay  for  a  perpetual  lease  of  $756.40  a 
year,  to  secure  8%  interest?  $9455. 

3.  Ground   rents  on  perpetual  lease,  yield  an  income  of 
$15642.90  a  year:  what  is  the  present  value  of  the  estate, 
allowing  1%  interest?  $223470. 

4.  What  is  the  initial  value  of  a  perpetual  leasehold  of 
$1600  a  year,  payable  semi-annually,  allowing  5^  interest, 
payable  annually?  $32400. 

SUGGESTION. — Here  the  yearly  payment  is  $1620,  by  allowing  5<& 
interest  on  the  half-yearly  payment  first  made. 

5.  What  is   the  initial  value  of  a  perpetual  leasehold  of 
$2500  a  year,  payable  quarterly,  interest  6%  payable  semi- 
annually;  6%  payable  annually:  6%  payable  quarterly? 

$41979.16f ;  $42604. 16|;  $41666.66| 


CASE     II. 

341.  To  find  the  present  value  of  a  deferred  per- 
petuity, when  the  payment,  the  interval,  the  rate, 
and  the  time  the  perpetuity  is  deferred  are  known. 

PROBLEM. — Find  the  present  value  of  a  perpetuity  of 
$250  a  year,  deferred  8  yr.,  allowing  6%  interest. 

SOLUTION. — Initial  value  of  perpetuity  of  $250  a  year,  by  last 
rule  =  $4166.66|  The  present  value  of  $4166.66f ,  due  8  yr.  hence, 
at  6/0  compound  interest,  =  $4166.66§  -f- 1.5938481  (Art.  335  ).  Use 
the  contracted  method,  reserving  3  decimal  places ;  the  quotient, 
$2614.22,  is  the  present  value  of  the  perpetuity. 

Rule. — Find  the  initial  value  of  the  perpetuity  by  the  last 
rule ;  then  find  the  present  worth  of  this  sum  for  the  time  the 


ANNUITIES.  311 

perpetuity  is  deferred,  by  Case  IV  of  Compound  Interest;  this 
wilt  be  the  present  value  required. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  present  value  of  a  perpetuity  of  $780  a  year, 
to  commence  in  12  yr.,  int.  5%.  $8686.66 

2.  Of  a  perpetual   lease  of  $160  a  year,  to  commence  in 
Syr.  4  mo.,  int.  1%.  $1823.28 

3.  Of  the  reversion  of  a  perpetuity  of  $540  a  year,  de* 
ferred  10  yr.,  int.  6%.  $5025.55 

4.  Of  an  estate  which,  in  5  yr. ,  is  to  pay  $325  a  year  for- 
ever: int.  8%,  payable  semi-annually.  $2690.67 

5.  Of  a  perpetuity  of  $1000  a  year,  payable  quarterly,  to 
commence  in  9  yr.  10  mo.  18  da.,  int.  10^,  payable  semi- 
annually.  $3858.88 

CASE   III. 

342.  Given  the  rate,  the  payment,  the  interval 
and  the  time  to  run,  to  find  the  present  value  of  an 
annuity  certain. 

PROBLEM. — 1.  Find  the  present  value  of  an  immediate 
annuity  of  $250  continuing  8  years,  6%  interest. 

SOLUTION. 

Present  value  of  immediate  perpetuity  of  $250,     .     .     .  =  $4166.67 

Present  value  of  perpetuity  of  $250,  deferred  8  yr.,  .     .  =    2614.22 

Pres.  val.  of  immediate  annuity  of  $250,  running  8  yr.,  =  $1552.45 

PROBLEM. — 2.  The  present  value  of  an  annuity  of  $680, 
to  commence  in  7  yr.  and  continue  10  yr.,  5%  int. 

SOLUTION. 

Pres.  val.  of  perpetuity  of  $680,  deferred  7  yr.,  at  5^,,  =$9665.27 
Pres.  val.  of  perpetuity  of  $680,  deferred  17  yr.,  at  5f0,  =  5933.64 
Pres.  val.  of  annuity  of  $680,  deferred  7  yr.,  to  run  10  yr.  =$3731.63 


312  RA  Y'S  HIGHER  ARITHMETIC. 

Rule.— Find  the  present  value  of  two  perpetuities  having 
the  given  rate,  payment,  and  interval,  one  of  them  commencing 
when  the  annuity  commences,  and  the  other  when  the  annuity 
ends.  The  difference  between  these  values  will  be  the  present 
value  of  the  annuity. 

NOTES.— 1.  This  rule  applies  whether  the  annuity  is  immediate  or 
deferred ;  in  the  latter,  the  time  the  annuity  is  deferred  must  be 
known,  and  used  in  getting  the  values  of  the  perpetuities. 

2.  By  using  the  initial  instead  of  the  present  values  of  the  per- 
petuities, the  rule  gives  the  initial  value  of  the  deferred  annuity, 
which  may  be  used  in  finding  its  final  or  forborne  value.  (Kern.  1, 
Case  IV.) 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  present  value  of  an  annuity  of  $125,  to  com- 
mence in  12  yr.  and  run  12  yr.,  int.  1ft.  $440.83 

2.  The  present  value  of  an  immediate  annuity  of  $400, 
running  15  yr.  6  mo.,  int.  8%.  $3484.41 

3.  The   present  value  of  an  annuity  of  $826.50,  to  com- 
mence in   3   yr.  and  run   13  yr.  9  mo.,    int.  6^,    payable 
semi-annually.  $6324. 69 

4.  The  present  value  of  an  annuity  of  $60,  deferred  12  yr. 
and  to  run  9  yr.,  int.  4±fo.  $257.17 

5.  Sold  a  lease  of  $480  a  year,  payable  quarterly,  having 
8  yr.  9  mo.  to  run,  for  $2500:  do  I  gain  or  lose,  int.  8%, 
payable  semi-annually?  Lose  $509.96 

CASE    IV. 

343.  Given  the  payment,  the  interval,  the  rate,  and 
time  to  run,  to  find  the  final  or  forborne  value  of  an 
annuity. 

PROBLEM. — Find  the  final  or  forborne  value  of  an  annuity 
of  $250,  continuing  8  yr.,  int.  6^. 


ANNUITIES.  313 

SOLUTION. — The  initial  value  of  a  perpetuity  of  $250,  at  6^,  = 
$4166.66§;  its  compound  interest,  at  6$  for  8  yr.,  =  $4166.6Cf  X 
.5938481  =  $2474.37,  the  final  or  forborne  value  of  the  annuity. 

Rule. — Consider  the  annuity  a  perpetuity,  and  find  its 
initial  value  by  Case  I.  The  compound  interest  of  this  sum, 
at  the  given  rate  for  the  time  the  annuity  runs,  will  be  the 
final  or  forborne  value. 

NOTES. — 1.  The  final  or  forborne  value  of  an  annuity  may  be 
obtained  by  finding  first  the  initial  value,  as  in  Case  III,  and  then 
the  compound  amount  for  the  time  the  annuity  runs. 

2.  The  present  value  of  an  annuity  can  be  obtained  by  finding 
first  the  forborne  value,  as  in  this  case,  and  then  the  present  worth 
for  the  time  the  annuity  runs. 


EXAMPLES  FOR  PRACTICE. 

1.  Find   the   forborne  value  of  an   immediate  annuity  of 
$300,  running  18  yr.,  int.  5^.  $8439.72 

2.  A  pays  $25  a  year  for  tobacco:  how  much  better  off 
would  he  have  been  in  40  yr.  if  he  had  invested  it  at  10^ 
per  annum?  $11064.81 

3.  Find  the  forborne  value  of  an  annuity  of  $75,  to  com- 
mence in  14  yr.,  and  run  9  yr.,  int.  6^.  $861.85 

SUGGESTION. — The  14  yr.  is  not  used.. 

4.  A  pays  $5  a  year  for  a  newspaper:  if  invested  at  9%, 
what  will  his  subscription  have  produced  in  50  yr.? 

$4075.42 

5.  An  annuity,  at  simple  interest  6%,  in  14  yr.,  amounted 
to  $116.76  :    what   would  have   been  the  difference  had   it 
been  at  compound  interest  6%?  $9.33 

6.  A  boy  just  9  yr.  old,  deposits  $35  in  a  bank :  if  he 
deposit  the  same  each  year  hereafter,  and  receive  10%,  com- 
pound interest,  what  will  be  the  entire  amount  when  he  is 
of  age?  $858.29 

H.  A.  27. 


314 


RAY'S  HIGHER  ARITHMETIC. 


The  present  value  of  $1  per  annum  in  any  number  of  yearSj  not 
exceeding  fifty -five. 


Yrs. 

4  per  cent. 

6  per  cent. 

6  per  cent. 

'  7  per  cent. 

8  per  cent. 

10  per  cent. 

1 
2 
3 
4 
5 

.961538 
1.886095 
2.775091 
3.629895 
4.451822 

.952381 
1.859410 
2.723248 
3.545951 
4.329477 

.943396 
1.833393 
2.673012 
3.4651  06 
4.212364 

.934579 

1.808018 
2.624316 
3.387211 
4.100197 

.925926 
1.783265 
2.577097 
3.312127 
3.992710 

.909091 
1.735537 
2.48(5852 
3.169865 
3.790787 

6 
7 
8 
9 
10 

5.242137 

6.002055 
6.732745 
7.435332 
8.110896 

5.075692 
5.786373 
6.463213 
7.107822 
7.721735 

4.917324 

5.582381 
6.209794 
6.801692 
7.360087 

4.766540 
5.389289 
5.971299 
6.515232 
7.023582 

4.622880 
5.206370 
5.746(539 
6.246888 
6.710081 

4.355261 
4.868419 
5.334926 
5.759024 
6.144567 

11 
12 
13 
14 
15 

8.760477 
9.385074 
9.985648 
10.563123 
11.118387 

8.306414 
8.863252 
9.303573 
9.898641 
10.379658 

7.886S75 
8.383844 
8.852683 
9.294984 
9.712249 

7.498674 
7.942686 
8.357651 
8.745468 
9.107914 

7.138964 
7.536078 
7.903776 
8.244237 
8.559479 

6.495061 
6.813692 
7.103356 
7.366687 
7.606080 

16 
17 
18 
19 
20 

11.652296 
12.165669 
12.659297 
13,133939 
13.590326 

10.837770 
11.274066 
11.689587 
12.085321 
12.462210 

10.105895 
10.477260 
10.827603 
11.158116 
11.469921 

9.446649 
9.763223 
10.059087 
10.335595 
10.594014 

8.851369 
9.121638 
9.371887 
9.603599 
9.818147 

7.823709 
8.021553 
8.201412 
8.3(54920 
8.513564 

21 
22 
23 
24 
25 

14.029160 
14.451115 
14.856842 
15.246963 
15.622080 

12.821153 
13.163003 
13.488574 
13.798642 
14.093945 

11.764077 
12.041582 
12.303379 
12.550358 
12.783356 

10.835527 
11.061241 
11.272187 
11.469334 
11.653583 

10.016803 
10.200744 
10.371059 
10.528758 
10.674776 

8.648694 
8.771540 
8.883218 
8.984744 
9.077040 

26 

27 
28 
29 
30 

15.982769 
16.329586 
16.663063 
16.983715 
17.292033 

14.375185 
14.643034 
14.898127 
15.141074 
15.372451 

13.003166 
13.210534 
13.406164 
13.590721 
13.764831 

11.825779 
11.986709 
12.137111 
12.277674 
12.409041 

10.809978 
10.935165 
11.051078 
11.158406 
11.257783 

9.160945 
9.237223 
9.306567 
9.369606 
9.426914 

31 
32 
33 
34 
35 

17.588494 
17.873552 
18.147646 
18.411198 
18.664613 

15.592811 
15.802677 
16.002549 
16.192904 
16.374194 

13.929086 
14.084043 
14.230230 
14.368141 
14.498246 

12.531814 
12.646555 
12.753790 
12.854009 
12.947672 

11.349799 
11.434999 
11.513888 
11  .586934 
11.654568 

9.479013 
9.526376 
9.569432 
9.608575 
9.644159 

36 
37 
38 
39 
40 

18.908282 
19.142579 
19.367864 
19.584485 
19.792774 

16.546852 
16.711287 
16.867893 
17.017041 
17.159086 

14.620987 
14.736780 
14.846019 
14.949075 
15.046297 

13.035208 
13.117017 
13.193473 
13.264928 
13.331709 

11.717193 
11.775179 

11.828869 
11.878582 
11.924613 

9.676508 
9.705917 
9.732651 
9.756956 
9.779051 

41 
42 
43 
44 
45 

19.993052 
20.185627 
20  370795 
20.548841 
20.720040 

17.294368 
17.423208 
17.545912 
17.662773 
17.774070 

15.138016 
15.224543 
15.306173 
15.383182 
15.455832 

13.394120 
13.452449 
13.506962 
13.557908 
13.605522 

11.967235 

12.006699 
12.043240 
12.077074 
12.108402 

9.799137 
9.817397 
9.833998 
9.849089 
9.862808 

46 
47 
48 
49 
50 

20.884654 
21.042936 
21.195131 
21.341472 
21.482185 

17.880067 
17.981016 
18.077158 
18.168722 
18.255925 

15.524370 
15.589028 
15.650027 
15.707572 
15.761861 

13.650020 
13.691608 
13.730474 
13.766799 
13.800746 

12.137409 
12.164267 
12.189136 
12.212163 
12.233485 

9.875280 
9.886618 
9.896926 
9.906296 
9.914814 

51 

52 
53 
54 
55 

21.617485 
21.747582 
21.872675 
21.992957 
22.108612 

18.338977 
18.418073 
18.493403 
18.565146 
18.633472 

15.813076 
15.861393 
15.906974 
15.949976 
15.990543 

13.832473 
13.862124 
13.889836 
13.915735 
13.939939 

12.253227 
12.271506 
12.288432 
12.304103 
12.318614 

9.92255* 
9.929599 
9.935999 
9.941817 
9.947107 

ANNUITIES.  315 


CALCULATIONS  BY  TABLE. 

344.  By  the    table  on   page   314,  some  interesting  and 
important  cases  in  annuities  can  be  solved,  among  which  are 
the  following  three  : 

CASE  v. 

345.  Given  the  rate,  time   to  run,  and  the  present 
or  final  value  of  an  annuity,  to  find  the  payment. 

PROBLEM. — An  immediate  annuity  running  11  yr.,  can  be 
purchased  for  $6000,:  what  is  the  payment,  int.  6%? 

SOLUTION. — The  present  value  of  an  immediate  annuity  of  $1  for 
11  yr.,  at  6^,,  is  $7.886875;  $6000  divided  by  this,  gives  $760.76,  the 
payment  required. 

» 

Rule. — Assume  $1  for  the  payment;  determine  the  present 
or  final  value  on  this  supposition,  and  divide  the  given  present 
or  final  value  by  it. 


EXAMPLES  FOR  PRACTICE. 

1.  How  much  a  year  should  I  pay,  to  secure  $15000  at 
the  end  of  17  yr.,  int.  7^?  $486.38 

2.  What   is  the  payment  of  an  annuity,  deferred  4  yr., 
running  16  yr.,  and  worth  $4800,  int.  6%?  $599.64 

CASE    VI. 

346.     Given  the  payment,  the  rate,  and  present  value 
of  an  annuity,  to  find  the  time  it  runs. 

PROBLEM. — In  what  time  will  a  debt  of  $10000,  drawing 
interest  at  6%,  be  paid  by  installments  of  $1000  a  year? 

SOLUTION. — The  $10000  may  be  considered  the  present  value  of 
an  annuity  of  $1000  a  year  at  6^;  but  $10000 -r- 1000— $10,   the 


316  RAY'S  HIGHER  ARITHMETIC. 

m 

present  value  of  an  annuity  of  $1  for  the  same  time  and  rate ;  by 
reference  to  the  table,  the  time  corresponding  to  this  present  value, 
under  the  head  of  6^,  is  15  yr.;  the  balance  then  due  may  be  thus 
found : 

Comp.  amt.  of  $10000  for  15  yr,,  at  6fc  (Art.  335),     .       =  $23965.58 

Final  val.  of  annuity  $1000  for  15  yr.,  at  6^,  (Art.  343),  =    23275.97 

Balance  due  at  end  of  15  yr $689.61 

Rule. — Divide  the  present  value  by  the  payment,  and  look 
in  the  table,  under  the  given  rate,  for  the  quotient;  the  number 
of  years  corresponding  to  the  quotient  or  to  the  tabular  number 
next  less,  will  be  the  number  of  full  intervals  required. 

NOTE. — The  difference  between  the  compound  amount  of  the  debt, 
and  the  forborne  value  of  the  annuity,  for  that  number  of  intervals, 
will  be  the  unpaid  balance. 


EXAMPLES  FOR  PRACTICE. 

1.  In  how  many  years  can  a  debt  of  $1000000,  drawing 
interest  at  6%,  be  discharged  by  a  sinking  fund  of  $80000  a 
year  ?  23  yr.,  and  $60083.43  then  unpaid. 

2.  In  how  many  years  can  a  debt  of  $30000000,  drawing 
interest  at  5^,  be  paid  by  a  sinking  fund  of  $2000000? 

28  yr.,  and  $798709.00  unpaid. 

3.  In  how  many  years  can  a  debt  of  $22000,  drawing  7^ 
interest,  be  discharged  by  a  sinking  fund  of  $2500  a  year  ? 

14  yr.,  and  $351.53  then  unpaid. 

4.  Let  the  conditions  be  the  same  as   those  of  the  illus- 
trative example,  and  let  each  $1000  payment    be   itself  a 
year's  accumulation  of  simple  interest :  what  would  be  the 
whole  time  required  to  discharge  the  debt? 

15  yr.  8  mo.  19  da. 

5.  Suppose  the  national  debt  $2000000000,  and  funded  at 
4%:  how  many  years  would  be  required  to  pay  it  off,  by  a 
sinking  fund  of  $100000000  a  year? 

41  yr.,  and  $3469275  unpaid. 


CONTINGENT  ANNUITIES.  317 


CASE    VII. 

347.  Given  the  payment,  time  to  run,  and  present 
value  of  an  annuity,  to  find  the  rate  of  interest. 

Rule. — Divide  the  present  value  by  the  payment ;  the  quotient 
will  be  the  present  value  of  $1  for  the  given  time  and  rate; 
look  in  tJie  table  and  opposite  the  given  number  of  years  for 
the  quotient  or  the  tabular  number  of  nearest  value,  and  at  Hie 
head  of  the  column  will  be  found  the  rate,  or  a  number  as 
near  the  true  rate  as  the  table  can  exhibit. 


EXAMPLES  FOR  PRACTICE. 

1.  If  $9000  is  paid  for  an  immediate  annuity  of  $750,  to 
run  20  yr.,  what  is  the  rate?  About  &%%. 

2.  If  an  immediate  annuity  of  $80,  running  14  yr.,  sells 
for  $650,  what  is  the  rate? 


CONTINGENT   ANNUITIES. 
DEFINITIONS. 

348.     1.  Contingent  Annuities  comprise  Life  Annuities, 
Dowers,  Pensions,  etc. 

2.  The  value  of  such  annuities  depends  upon  the  expecta- 
tion of  life. 

3.  Expectation  of  Life  is  the  average  number  of  years 
that  a  person  of  any  age  may  be  expected  to  live. 

4.  Tables,  called  "Mortality  Tables,"  have  been  prepared 
in  England  and  in  this  country  for  the  purpose  of  ascertain- 
ing how  many  persons  of  a  given  number  and  of  a  certain 


318 


RAY'S  HIGHER  ARITHMETIC. 


age  would  die  during  any  one  year,  and  in  how  many  years 
the  whole  number  would  die. 

KEMARK. — These  tables,  though  not  absolutely  accurate,  are 
based  upon  so  large  a  number  of  observations  that  their  approx- 
imation is  very  close.  Legal,  medical,  and  scientific  authorities 
use  them  in  discussing  vital  statistics,  and  insurance  companies 
make  them  a  basis  for  the  transaction  of  business. 

349.  The  following  table  differs  but  slightly  from  those 
prepared  in  this  country: 

CARLISLE  TABLE 

Of  Mortality,  based   upon  Obsemations  at  Carlisle  (Eng.),  shmving  the 
Rate  of  Extinction  of  10,000  lives. 


i 

•< 

Number  of 
Survivors. 

Number  of 
Deaths. 

| 

< 

Number  of 
Survivors. 

Number  of 
Deaths. 

o 
b£ 
*% 

Number  of 
Survivors. 

Number  of 
Deaths. 

0 

10000 

1539 

35 

5362 

55 

70 

2401 

124 

1 

8461 

682 

36 

5307 

56 

71 

2277 

134 

2 

7779 

505 

37 

5251 

57 

72 

2143 

146 

3 

7274 

276 

38 

5194 

58 

73 

1997 

156 

4 

6998 

201 

39 

5136 

62 

74 

1841 

166 

5 

6797 

121 

40 

5075 

66 

75 

1675 

160 

6 

6676 

82 

41 

5009 

69 

76 

1515 

156 

7 

6594 

58 

42 

4940 

71 

77 

1359 

146 

8 

6536 

43 

43 

4869 

71 

78 

1213 

132 

9 

6493 

33 

44 

4798 

71 

79 

1081 

128 

10 

6460 

29 

45 

4727 

70 

80 

953 

116 

11 

6431 

31 

46 

4657 

69 

81  • 

837 

112 

12 

6400 

32 

47 

4588 

67 

82 

725 

102 

13 

6368 

33 

48 

4521 

63 

83 

623 

94 

14 

6335 

35 

49 

4458 

61 

84 

529 

84 

15 

6300 

39 

1  50 

4397 

59 

85 

445 

78 

16 

6261 

42 

1  51 

4338 

62 

86 

367 

71 

17 

6219 

43 

52 

4276 

65 

87 

296 

64 

18 

6176 

43 

53 

4211 

68 

88 

232 

51 

19 

6133 

43 

54 

4143 

70 

89 

181 

39 

;  20 

6090 

43 

i  55 

4073 

73 

90 

142 

37 

1  21 

6047 

42 

56 

4000 

76 

91 

105 

30 

22 

6005 

42 

57 

3924 

82 

92 

75 

21 

23 

5963 

42 

58 

3842 

93 

93 

54 

14 

24 

5921 

42 

59 

3749 

106 

94 

40 

10 

25 

5879 

43 

60 

3643 

122 

95 

30 

7 

26 

5836 

43 

61 

3521 

126 

96 

23 

5 

27 

5793 

45 

62 

3395 

127 

97 

18 

4 

28 

5748 

50 

63 

3268 

125 

98 

14 

3 

29 

5698 

56 

64 

3143 

125 

99 

11 

2 

30 

5642 

57 

65 

3018 

124 

100 

9 

2 

31 

5585 

57 

66 

2894 

123 

101 

7 

2 

32 

5528 

56 

67 

2771 

123 

102 

5 

2 

83 

5472 

55 

68 

2648 

123 

103 

3 

2 

34 

5417 

55 

69 

2525 

124 

104 

1 

1 

CONTINGENT  ANNUITIES. 


319 


TABLE 

Showing  the  values  of  Annuities  on  Single  Lives,  according  to  the 
Carlisle  Table  of  Mortality. 


Age. 

4  per  ct. 

5  per  ct. 

6  per  ct. 

7  per  ct. 

Age. 

4  per  ct. 

5  per  ct. 

6  per  ct. 

7  per  ct. 

0 

14.28164 

12.083 

10.439 

9.177 

52 

12.25793 

11.154 

10.208 

9.392 

1 

16.55455 

13.995 

12.078 

10.605 

53 

11.94503 

10.892 

9.988 

9.205 

2 

17.72616 

14.983 

12.925 

11.342 

54 

11.62673 

10.624 

9.761 

9.011 

3 

18.71508 

15.824 

13.652 

11.978 

55 

11.29961 

10.347 

9.524 

8.807 

4 

19.23133 

16.271 

14.042 

12.322 

56 

10.96607 

10.063 

9.280 

8.595 

5 

19.59203 

16.590 

14.325 

12.574 

57 

10.62559 

9.771 

9.027 

8.375 

6 

19.74502 

16.735 

14.460 

12.698 

58 

10.28647 

9.478 

8.772 

8.153 

7 

19.79019 

16.790 

14.518 

12.756 

59 

9.96331 

9.199 

8.529 

7.940 

8 

19.76443 

16.786 

14.526 

12.770 

60 

9.66333 

8.940 

8.304 

7.743 

9 

19.69114 

16.742 

14.500 

12.754 

61 

9.39809 

8.712 

8.108 

7.572 

10 

J  9.58339 

16.669 

14.448 

12.717 

62 

9.13676 

8.487 

7.913 

7.403 

11 

19.45857 

16.581 

14.384 

12.669 

63 

8.87150 

8.258 

7.714 

7.229 

12 

19.33493 

16.494 

14.321 

12.621 

64 

8.59330 

8.016 

7.502 

7.042 

13 

19.20937 

16.406 

14.257 

12.572 

65 

8.30719 

7.765 

7.281 

6.847 

14 

19.08182 

16.316 

14.191 

12.522 

66 

8.00966 

7.503 

7.049 

6.641 

15 

18.95534 

16.227 

14.126 

12.473 

67 

7.69980 

7.227 

6.803 

6.421 

16 

18.83(536 

16.144 

14.067 

12.429 

68 

7.37976 

6.941 

6.546 

6.189 

17 

18.72111 

16.066 

14.012 

12.389 

69 

7.04881 

6.643 

6.277 

5.945 

18 

18.60656 

15.987 

13.956 

12.348 

70 

6.70936 

6.336 

5.998 

5.690 

19 

18.48649 

15.904 

13.897 

12.305 

71 

6.35773 

6.015 

5.704 

5.420 

20 

18.36170 

15.817 

13.835 

12.259 

72 

6.02548 

5.711 

5.424 

5.162 

21 

18.23196 

15.726 

13.769 

12.210 

73 

5.72465 

5.435 

5.170 

4.927 

22 

18.09386 

15.628 

13.697 

12.156 

74 

5.45812 

5.190 

4.944 

4.719 

23 

17.95016 

15.525 

13.621 

12.098 

75 

5.23901 

4.989 

4.760 

4.549 

24 

17.80058 

15.417 

13.541 

12.037 

76 

5.02399 

4.792 

4.579 

4.382 

25 

17.64486 

15.303 

13.456 

11.972 

77 

4.82473 

4.609 

4.410 

4.227 

26 

17.48586 

15.187 

13.368 

11.904 

78 

4.62106 

4.422 

4.238 

4.067 

27 

17.32023 

15.065 

13.275 

11.832 

79 

4.39345 

4.210 

4.040 

3.883 

28 

17.15412 

14.942 

13.182 

11.759 

80 

4.1828!) 

4.015 

3.858 

3.713 

29 

16.99683 

14.827 

13.096 

11.693 

81 

3.95309 

3.799 

3.656 

3.523 

30 

16.85215 

14.723 

13.020 

11.636 

82 

3.74634 

3.606 

3.474 

3.352 

31 

16.70-51  1 

14.617 

12.942 

11.578 

83 

3.53409 

3.406 

3.286 

3.174 

32 

16.55246 

14.506 

12.860 

11.516 

84 

3.32856 

3.211 

3.102 

2.999 

33 

16.39072 

14.387 

12.771 

11.448 

85 

3.11515 

3.009 

2.909 

2.815 

34 

16.21943 

14.260 

12.675 

11.374 

86 

2.92831 

2.830 

2.739 

2.652 

35 

16.04123 

14.127 

12.573 

H.295 

87 

2.77593 

2.685 

2.599 

2.519 

36 

15.85577 

13.987 

12.465 

11.211 

88 

2.68337 

2.597 

2.515 

2.439 

37 

15.66586 

13.843 

12.354 

11.124 

89 

2.57704 

2.495 

2.417 

2.344 

38 

15.47129 

13.695 

12.239 

11.033 

90 

241621 

2.339 

2.266 

2.198 

39 

15.27184 

13.542 

12.120 

10.939 

91 

2.39835 

2.321 

2.248 

2.180 

40 

15.07363 

13.390 

12.002 

10.845 

92 

2.49199 

2.412 

2.337 

2.266 

41 

14.88314 

13.245 

11.890 

10.757 

93 

2.59955 

2.518 

2.440 

2.367 

42 

14.69466 

13.101 

11.779 

10.671 

94 

2.64976 

2.569 

2.492 

2.419 

4:j 

14.50529 

12.957 

11.668 

10.585 

95 

2.67433 

2.596 

2.522 

2.451 

44 

14.30874 

12.806 

11.551 

10.494 

96 

2.62779 

2.555 

2.486 

2.420 

45 

14.10460 

12.648 

11.428 

10.397 

97 

2.49204 

2.428 

2.368 

2.309 

4t> 

13.88928 

12.480 

11.296 

10.292 

98 

2.33222 

2.278 

2.227 

2.177 

47 

13.6(5208 

12.301 

11.154 

10.178 

99 

2.08700 

2.045 

2.004 

1.964 

48 

1:5.41914 

12.107 

10.998 

10.052 

100 

1.65282 

1.624 

1.596 

1.569 

49 

13.15312 

11.892 

10.823 

9.908 

101 

1.210U5 

1.192 

1.175 

1.159 

50 

12.86902 

11.660 

10.631 

9.749 

102 

0.76183 

0.753 

0.744 

0.735 

51 

12.56581 

11.410 

10.422 

9.573 

103 

0.33051 

0.317 

0.314 

0.312 

320  RAY'S  HIGHER  ARITHMETIC. 


CALCULATIONS  BY  TABLE. 

350.  The  preceding    table   of  life  annuities    shows    the 
sura  to  be  paid  by  a  person  of  any  age,  to  secure  an  an- 
nuity of  $1  during  the  life  of  the  annuitant. 

CASE    I. 

351.  To   find   the  value  of  a  given  annuity  on  the 
life  of  a  person  whose  age  is  known. 

Rule. — Find  from  the  table  the  value  of  a  life  annuity  of 
$1,  for  the  given  age  and  rate  of  interest,  and  multiply  it  by 
the  payment  of  the  given  annuity. 

REMARKS. — 1.  To  find  the  value  of  a  life-estate  or  widow's  dower 
(which  is  a  life-estate  in  one  third  of  her  husband's  real  estate): 
Estimate  the  value  of  the  property  in  which  the  life-estate  is  held  ;  the  yearly 
interest  of  this  sum,  at  an  ac/reed  rate,  will  be  a  life-annuity,  whose  value  for 
the  given  aye  and  rate  will  be  the  value  of  the  life-estate. 

2  The  reversion  of  a  life-annuity,  life-estate,  or  dower  is  found 
by  deducting  its  value  from  the  value  of  the  property. 


EXAMPLES  FOR  PRACTICE. 

1.  What  must  be  paid  for  a  life-annuity  of  $650  a  year, 
by  a  person  aged  72  yr.,  int.  7%?  $3355.30 

2.  What  is  the  life-estate  and  reversion  in  $25000,  age 
55  yr.,  int.  6%?  Life-estate,  $14286;  rev.,  $10714. 

3.  The  dower  and  reversion  in  $46250,  age  21   yr.,  int. 
6%?  Dower,  $12736.33;  rev.,  $2680.34 

CASE    II. 

352.  To  find  how  large  a  life-annuity  can  be  pur- 
chased for  a  given  sum,  by  a  person  whose  age  is 
known. 


CONTINGENT  ANNUITIES.  321 

Rule. — Assume  $1  a  year  for  tlie  annuity;  find  from  tlie 
table  its  value  for  the  given  age  and  rate  of  interest,  and  divide 
the  given  cost  by  it;  the  quotient  will  be  the  payment  required. 


EXAMPLES  FOR  PRACTICE. 

How  large  an  annuity  can  be  purchased : 

1.  For  $500,  age  26  yr.,  int.  6^?-  $37.40 

2.  For  $1200,  age  43,  int.  5^?  $92.61 

3.  For  $840,  age  58,  int.  Ifi't  $103.03 

CASE    III. 

353.  To  find  the  present  value  of  the  reversion 
of  a  given  annuity ;  that  is,  what  remains  of  it,  after 
the  death  of  its  possessor,  whose  age  is  known. 

Rule. — Find  the  present  value  of  the  annuity  during  its 
whole  continuance;  find  its  value  during  the  given  life;  their 
difference  will  be  the  value  of  the  reversion. 

NOTE. — It  will  save  work,  to  consider  the  annuity  as  $1  a  year,  then 
apply  the  rule,  using  the  tables  in  Art,  335  and  Art.  350,  and  multiply 
the  result  by  the  given  payment. 


EXAMPLES  FOR  PRACTICE. 

1 .  Find  the  present  value  of  the  reversion  of  a  perpetuity 
of  $500  a  year,  after  the  death  of  a  person  aged  47,  int. 
5%.  $3849.50 

2.  Of  the  reversion  of  an  annuity  of  $165  a  year  for  30 
yr.,  after  the  death  of  a  person  38  yr.  old,  int.  6%. 

$251.76 

3.  Of  the  reversion  of  a  lease  of  $1600  a  year,  for  40  yr., 
after  the  death  of  A,  aged  62,  -int.  1%.  $9485.93 


322  RA  YJ  S  HIGHER  ARITHMETIC. 

PERSONAL   INSURANCE. 
DEFINITIONS. 

354.  1.  Personal  Insurance  is  of  two  kinds:  (1.)  Life 
bisurance;  (2.)  Accident  Insurance. 

2.  Life    Insurance   is  a  contract   in   which   a  company 
agrees,  in  consideration  of  certain  premiums  received,  to  pay 
a  certain  sum  to  the  heirs  or  assigns  of  the  insured  at  his 
death,  or  to  himself  if  he  attains  a  certain  age. 

3.  Accident    Insurance    is    indemnity    against    loss    by 
accidents. 

4.  The  Policies  issued  by  life  insurance  companies  are: 
(1.)   Term  Policies;    (2.)   Ordinary  Life    Policies;    (3.)  Joint 
Life  Policies',  (4.)  Endowment  Policies;  (5.)  Reserved  Endow- 
ment Policies;    (6.)   Tontine  Savings  Fund  Policies. 

5.  The  chief  policies  are,  however,  the  Ordinary  Life  and 
the  Endowment. 

6.  The  Ordinary  Life  Policy  secures  a  certain  sum  of 
money  at  the  death  of  the  insured.     Premiums  may  be  paid 
annually  for  life,  semi-annually,  quarterly,  or  in   one  pay- 
ment in  advance;  or  the  premiums  may  be  paid  in  5,  10,  15, 
or  20  annual  payments. 

7.  An  Endowment  Policy  secures  to  the  person  insured 
a  certain  sum  of  money  at  a  specified  time,  or  to  his  heirs  or 
assigns  if  he  die  before  that  time. 

REMARK. — It  will  be  advantageous  for  the  student  to  examine 
an  "  application  "  and  a  "policy"  taken  from  some  case  of  actual 
insurance;  by  a  short  study  of  such  papers,  the  nature  of  the  insur- 
ance contract  will  be  learned  more  easily  than  by  any  mere  verbal 
description ;  additional  light  may  be  had  from  the  reports  pub- 
lished by  various  companies. 

355.  1.  The  following  is  a  condensed  table  of  one  of  the 
leading  companies: 


PERSONAL  INSURANCE. 


TABLE. 

Annual  Premium  Rates  for  an  Insurance  of  $1000. 


LIFE  POLICIES. 

ENDOWMENT  POLICIES. 

Payable  at  death  only. 

Payable  as  indicated,  or  at  death,  if  prior. 

Age. 

Annual  Payments. 

Single 
Payment 

Age. 

In 

10  years 

In 
15  years 

In 

20  years 

For  life 

20  years 

10  years 

20  to 

20  to 

25 

$19  89 

$27  39 

$42  56 

$326  58 

25 

$103  91 

$66  02 

$47  68 

26 

20  40 

27  93 

43  37 

332  58 

26 

104  03 

66  15 

47  82 

27 

20  93 

28  50 

44  22 

338  83 

27 

104  16 

66  29 

47  98 

28 

21  48 

29  09 

45  10 

345  31 

28 

104  29 

66  44 

48  15 

29 

22  07 

29  71 

46  02 

352  05 

29 

104  43 

66  60 

48  33 

30 

22  70 

30  36 

46  97 

359  05 

30 

104  58 

66  77 

48  53 

31 

23  35 

31  03 

47  98 

366  33 

31 

104  75 

66  96 

48  74 

32 

24  05 

31  74 

49  02 

373  89 

32 

104  92 

67  16 

48  97 

33 

24  78 

3248 

50  10 

381  73 

33 

105  11 

67  36 

49  22 

34 

25  56 

33  26 

51  22 

389  88 

34 

105  31 

67  60 

49  49 

35 

26  38 

34  08 

52  40 

398  34 

35 

105  53 

67  85 

49  79 

36 

27  25 

34  93 

53  63 

407  11 

36 

105  75 

68  12 

50  11 

37 

28  17 

35  83 

54  91 

416  21 

37 

106  00 

68  41 

50  47 

38 

29  15 

36  78 

56  24 

425  64 

38 

106  28 

68  73 

50  86 

39 

30  19 

37  78 

57  63 

435  42 

39 

106  58 

69  09 

51  30 

40 

31  30 

38  83 

59  09 

445  55 

40 

106  90 

69  49 

51  78 

41 

32  47 

39  93 

60  60 

456  04 

41 

107  26 

69  92 

52  31 

42 

33  72 

41  10 

62  19 

466  89 

42 

107  65 

70  40 

52  89 

43 

35  05 

42  34 

63  84 

478  11 

43 

108  08 

70  92 

53  54 

44 

36  46 

43  64 

65  57 

489  71 

44 

108  55 

71  50 

54  25 

45 

37  97 

45  03 

67  37 

501  69 

45 

109  07 

72  14 

55  04 

46 

39  58 

46  50 

69  26 

514  04 

46 

109  65 

72  86 

55  91 

47 

41  30 

48  07 

71  25 

526  78 

47 

110  30 

73  66 

56  89 

48 

43  13 

49  73 

73  32 

53988 

48 

111  01 

74  54 

57  96 

49 

45  09 

51  50 

75  49 

553  33 

49 

111  81 

75  51 

59  15 

50 

47  18 

53  38 

77  77 

567  13 

50 

112  68 

76  59 

60  45 

51 

49  40 

55  38 

80  14 

581  24 

51 

113  64 

77  77 

61  90 

52 

51  78 

57  51 

82  63 

595  66 

52 

114  70 

79  07 

63  48 

53 

54  31 

59  79 

85  22 

610  36 

53 

115  86 

80  51 

65  22 

54 

57  02 

62  22 

87  94 

625  33 

54 

117  14 

82  09 

67  14 

55 

59  91 

64  82 

90  79 

640  54 

55 

118  54 

83  82 

69  24 

56 

63  00 

67  60 

93  78 

655  99 

56 

120  09 

85  73 

57 

66  29 

70  59 

96  91 

671  64 

57 

121  78 

87  84 

58 

69  82 

73  78 

100  21 

687  48 

58 

123  64 

90  15 

59 

73  60 

77  22 

103  68 

703  49 

59 

125  70 

92  70 

60 

77  63 

80  91 

107  35 

719  65 

60 

127  96 

95  50 

61 

81  96 

84  88 

111  23 

735  92 

61 

130  45 

62 

86  58 

89  16 

115  32 

752  26 

62 

133  19 

63 

91  54 

93  7.6 

119  66 

768  67 

63 

136  20 

64 

96  86 

98  73 

124  28 

785  10 

64 

139  52 

65 

102  55 

104  10 

129  18 

801  52 

65 

143  16 

i 

324  HAY'S  HIGHER  ARITHMETIC. 

2.  Quantities  considered  in  Life  Insurance  are: 
'  1.  Premium  on^lOOO. 

2.  The  Gain  or  Loss. 

3.  Amount  of  the  Policy. 

4.  Age  of  the  Insured. 

1 5.  The  Term  of  years  of  Insurance. 
These  quantities  give  rise  to  five  classes  of  problems, 
but  they  involve  no  new  principles,  and  by  the  aid  of  the 
preceding  tables  they  are  easily  solved.  Simple  interest  is 
intended  where  interest  is  mentioned  in  the  following  prob- 
lems : 

EXAMPLES  FOR  PRACTICE. 

1.  W.  R.  Hamilton,  aged  40  years,  took  a  life  policy  for 
$5000.     Required  the  annual  premium?  $156.50 

2.  Conditions   as  above,  how  much  would  he   have  paid 
out  in  premiums,  his  death  having  occurred   after  he  was 
53?  $2191. 

3.  Conditions  the  same,  what  did  the  premiums  amount 
to,  interest  6%?  $3045.49 

4.  James   Bragg,  aged   50  years,  took  out  an  endowment 
policy  for  $20000,  payable  in  10  years,  and  died  after  making 
6   payments :    how  much   less   would   he  have   paid   out  by 
taking  a  life  policy  for  the  same  amount,  the  premium  pay- 
able annually  ?  $7860. 

5.  Thomas  Winn,  28  years  of  age,  took   out   an    endow- 
ment policy  for  $10000,  payable  in  10  years;  he  died  in  18 
months:    what  was   the  gain,  interest  on   the   premiums  at 
6%,  and  how  much  greater  would  the  profit  have  been  had 
he  taken  a  life  policy,  premiums  payable  annually? 

$7789.052;  $1755.57,  profit. 

6.  P.    Darling   took  out  a  life  policy   at    the  age  of  40 
years,  and   died  just   after   making  the  tenth  payment ;  his 
premiums  amounted  to  $3975.10,  interest  6^;  required  the 
amount  of  his  policy?  $10000. 


TOPICAL   OUTLINE. 


325 


7.  R.  C.  Storey  took  out  an  endowment  policy  for  SI 0000 
for  15  years ;  he  lived  to  pay  all  of  the  premiums ;  but  had 
he  put  them,  instead,  at  6^  interest  as  they  fell  due,  they 
would  have  amounted  to  $15426. 78:  what  was  his  age? 

40  years. 

8.  Allen  Wentworih    had   his  life  insured  at  the  age  of 
twenty,  on  the  life  plan,  for  $8000,  premium  payable  annu- 
ally :  how  old   must  he  be,  that  the   sum  of  the  premiums 
may  exceed  the  policy?  71  years  old. 

9.  T.  B.  Bullene,  aged  40   years,  took  out  a  life  policy 
for  $30000,  payments   to  cease   in   5   years,  the  rate  being 
$9.919  on  the  $100  ;  his  death  occurred  two  months  after  he 
had  made  the  third   payment:  what  was  gained  over  and 
above  the  premiums,  interest  6%?  $20448. 

10.  F.  M.  Harrington  took  out  an  endowment  policy  for 
$11000  when  he  was    42;  at  its  maturity  he  had  paid  in 
premiums  $635.80   more  than  the  face  of  the  policy:  what 
was  the  period  of  the  endowment  ?  20  years. 


Topical  Outline. 
APPLICATIONS  OF  PERCENTAGE. 

(With  Time.) 

1.  Definitions:— Interest,  Principal,  Rate,  Amount,  Legal 
Rate,  Usury,   Notes  Promissory,   Face,  Payee,  In- 
dorser,  Demand  Note,  Time  Note,   Principal   and 
Surety,  Maturity,  Protest. 
(  Methods 
[M       or 
1.  Simple  Interest.  J  I  Rulcs 


2.  Five  Cases...  - 


Common. 
Aliquots. 

Six  and  Twelve  per  ct. 
I  Exact  Interest. 

2.  Rule. — Formula. 

3.  Rule.— Formula. 

4.  Rule. — Formula. 

5.  Rule. — Formula. 
^  3.  Annual  Interest. — Rule. 


326 


RAY'S  HIGHER  ARITHMETIC. 


Topical  Outline. — (Continued.} 
APPLICATIONS  OF  PEKCENTAGE. 

(With  Time.) 


2.  Partial  Payments.. 


1.  Definitions : — Payment,  Indorsement. 


2.  Rules.. 


3.  Discount... 


1.  True  Discount... 


2.  Bank  Discount. 


1.  U.   S.   Rule.— Principles.     Connecticut, 
'        Vermont,  and  Mercantile  Rules. 

1.  Definitions:— Present  Worth, Discount. . 

2.  Rule. 

1.  Definitions  :—Bank,     Deposit,    Issue, 

Check,  Drafts,  Drawee,  Payee,  In- 
dorsement, Discount,  Proceeds, 
Days  of  Grace,  Time  to  Ruir. 

2.  Four  Cases,  Rules. 


4.  Exchange-j 


1.  Definitions  :—  Domestic  and  Foreign  Exchange,  Bill,  Set,  Rate, 

Course,  Par,  Intrinsic,  and  Commercial. 

f  1.  Domestic. 

2.  Kinds  J  f  L  Direct    Tabie  of  Values. 

L2'Foreisn"  fl.  Definitions  :-ArbIt»- 

[  2.  Circular.  I 


tion,  Simple  and 
Compound. 


L 


2.  Rules. 


5.  Equation  of  Payments... 


6.  Settlement  of  Accounts. 


1.  Definitions:— Equated  Time,  Term  of  Credit, 
Average  Term,  Averaging  Account,  Clos- 
ing Account,  Focal  Date. 

2   Principles. 

3.  Rules. 


1.  Definitions. 


1.  Accounts  Current;  Rule. 
Account  Sales;  Rule. 
Storage. 


7.  Compound  Interest.  I  L  Definitions :- Comp.  Int.,  Comp.  Amt. 
I  2. 


8.  Annuities.. 


9.  Personal  Insurance. 


Four   Cases,  Rules. 

f  1.  Definitions:— Perpetual,  Limited,  Certain,  Contingent.  Im- 
mediate, Deferred,  Forborne,  Final  Value,  Initial  Value, 
Present  Value. 

I  2.  Seven  Cases.    Rules.    Table. 

1.  Definitions. 

2.  Table. 


XYIL    PAETWEESHIP. 

DEFINITIONS. 

356.  1.  Partnership  is  the  association  of  two  or  more 
persons  in  a  business  of  which  they  are  to  share  the  profits 
and  the  losses.     The  persons  associated  are  called  partners; 
together  they  constitute  a  Firm,  Company,  or  House. 

2.  The  Capital  is  the  money  employed  in  the  business ; 
the  Assets  or  Resources  of  a  firm   are  its   property,  and 
opposed  to  these  are  its  Liabilities  or  Debts. 

3.  Partnership  has  two  cases:  (1.)  When  all  the  shares 
of  the  capital   are   continued  through   the  same  time ;  (2.) 
When  the  full  shares  are   not   continued    through  the  same 
time.      The  first  is  called  Simple   Partnership;   the  second, 
Compound  Partnership. 

PRINCIPLE. — Gains  and  losses  are  shared  in  proportion  to 
the  sums  invested  and  the  periods  of  investment. 

CASE    I. 

357.  To  apportion  the  gain  or  loss,  when  all  of  each 
partner's  stock  is  employed  through  the  same  time. 

PROBLEM.— A,  B,  and  C  are  partners,  with  $3000,  $4000, 
and  $5000  stock,  respectively ;  if  they  gain  $5400,  what  is 
each  one's  share? 

OPERATION. 

SoLimoN.-The  whole  3  ^  of  $  5  4  0  0  =  $  1  3  5  0,  A's  share, 

stock  is  $12000,  of  which  4    4  «       54QO  —     1800    B's       " 

A   owns  T%,  B  A,  C.-&;  j^  _/T  «       5400=     2250*  C's       " 

hence,   by   the    principle  12    '  $5^00;  whole, 

stated,  A  should  have  ^ 

of  the  gain,  or  $1350;  in  like  manner,  B,  $1800;  C,  $2250. 

(327) 


328  RAY'S  HIGHER  ARITHMETIC. 

Rule. — Divide  the  gain  or  toss  among  the    partners  in  pro- 
portion to  their  shares  of  the  stock. 

REMARK. — The  division  may  be  made  by  analysis  or  by  simple 
proportion. 

EXAMPLES  FOR  PRACTICE. 

1.  A  and  B  gain  in  one  year  $3600;  their  store  expenses 
are  $1500.     If  A's  stock  is  $2500  and  B's  $1875,  how  much 
does  each  gain?  A  $1200,  B  $900. 

2.  A,  B,  and  C  are  partners ;   A  puts  in  $5000,  B  6400, 
C  $1600.     C  is  allowed  $1000  a  year  for  personal  attention 
to  the  business ;  their  store  expenses  for  one  year  are  $800, 
and   their  gain,   $7000.     Find   A's  and  B's  gain,  and   C's 
income.  A  $2000,  B  $2560,  C  $1640. 

3.  A,  B,  and  C  form  a  partnership ;  A  puts  in  $24000, 
B  $28000,  C  $32000;  they  "lose  £  of  their  stock  by  a  fire, 
but  sell  the  remainder  at  f  more  than  cost:  if  all  expenses 
are  $8000,  .what  is  the  gain  of  each  ? 

A  $5714.28£,  B  $6666.66|,  C  $7619.04if 

4.  A,  B,  and   C  are  partners;  A's  stock    is  $5760,   B's, 
$7200;  their  gain  is  $3920,  of  which  C  has  $1120:  what  is 
C's  stock,  and  A's  and  B's  gain  ? 

C's  stock,  $5184;  A's  gain,  $1244. 44£;  B's  gain,  $1555.55f 

5.  A,    B,   and   C    are  partners;  A's    stock,    $8000;  B's, 
$12800;  C's,  $15200;  A  and  B  together  gain  $1638   more 
than  C :  what  is  the  gain  of  each  ? 

A  $2340 ;  B  $3744  ;  C  $4446. 

6.  A,  B,  and  C  have  a  joint  capital  of  $27000;  none  of 
them    draw    from    the    firm,   and    when   they   quit   A   has 
$20000;    B,  $16000;  C,  $12000:  how  much  did  each  con- 
tribute? A,  $11250;  B,  $9000;  C,  $6750. 

7.  A,  B,  C,  and  D  gain  30%  on  the  stock  ;  A,  B,  and  C 
gain  $1150;    A,  B,  D,  $1650;    B,  C,  D,  $1000;    A,  C,  D, 
$1600  :  what  was  each  man's  stock? 

A,  $2666f ;  B,  $666|;  C,  $500;  D,  $2166|. 


PARTNERSHIP.  329 


CASE   II. 

358.  To  apportion  the  gain  or  loss  when  the  full 
shares  are  not  continued  through  the  same  period. 

PROBLEM. — A,  B,  and  C  are  partners;  A  puts  in  $2500 
for  8  mo.;  B,  $4000  for  6  mo.;  C,  $3200  for  10  mo.;  their 
net  gain  is  $4750 :  divide  the  gain. 

SOLUTION. — A's  OPERATION. 

capital     ($2500),  $2500X     8  =  $2  0000,  A's  equivalent, 

used  8  months,  is  $  4  0  0  0  X     6  =     2  4  0  0  0,  B's          " 

equivalent  to8X  $3200X10=    32000,  C's          " 

$2500,  or  $20000,  $  7  6  0  0  0 

used  1  month ;  B's  f  f  of  $  4  7  5  0  =  $  1  2  5  0,  A's  share, 

capital     (  $4000  ),  f£  of  $  4  7  5  0  =  $  1  5  0  0,  B's       « 

used  6  months,  is  W  of  $  4  7  5  0  =  $  2  0  0  0,  C*      •< 

equivalent  to  6  X 

$4000,  or  $24000,  used  1  month  ;  C's  capital  ($3200),  used  10  months, 
is  equivalent  to  10  X  $3200,  or  $32000  used  1  month.  Dividing  the 
gain  ($4750)  in  proportion  to  the  stock  equivalents,  $20000,  $24000, 
$32000,  used  for  the  same  time  (1  month),  the  results  will  be  the 
gain  of  each;  A's  $1250,  B's  $1500,  C's  $2000. 

Rule. — Multiply  each  partner's  stock  by  the  time  it  is  used; 
and  divide  the  gain  or  loss  in  proportion  to  the  products  so 
obtained. 

EXAMPLES  FOR  PRACTICE. 

1.  A  begins  business  with  $6000 ;  at  the  end  of  6  mo.  he 
takes  in  B,  with  $10000;  6  mo.  after,  their  gain  is  $3300: 
what  is  each  share?  A's,  $1800;  B's,  $1500. 

2.  A  and  B  are  partners ;  A's  stock  is  to  B's,  as  4  to  5 ; 
after  3  mo.,  A  withdraws  f  of  his,  and  B  f  of  his:  divide 
their  year's  gain,  $1675.  A,  $800  ;  B,  $875. 

3.  A.  B,  and  C  join  capitals,  which  are  as  y,  i>  i;  after 
4  mo.,  A  takes  out  |-  of  his ;  after  9  mo.  more,  their  gain  is 
$1988:  divide  it.  A,  $714;  B,  $728;  C,  $546. 

H    A.  28. 


330  RAY'S  HIGHER  ARITHMETIC. 

4.  A  and  B  are  partners;  A  puts  in   $2500;  B,  $1500; 
after  9  mo. ,  they  take  in  C  with  $5000 ;  9  mo.  after,  their 
gain  is  $3250 :  what  is  each  one's  gain  ? 

A's,  $1250;  B's,  $750;  C's,  $1250. 

5.  A  and  B  are  partners,  each  contributing  $1000 ;  after 
3  months,  A  withdraws  $400,  which  B  advances ;  the  same 
is  done  after  3  months  more ;  their  year's  gain  is  $800 :  what 
should  each  get  ?  A,  $200  ;  B,  $600. 

6.  A,  B,  and  C  are  employed   to  empty  a  cistern  by  two 
pumps  of  the  same  bore ;  A  and  B  go  to  work  first,  making 
37  and  40  strokes  respectively  a  minute ;  after  5  minutes, 
each   makes  5  strokes  less  a  minute ;    after  10  minutes,  A 
gives   way  to   C,  who  makes  30   strokes  a  minute  until  the 
cistern  is  emptied,  which  was  in  22  minutes  from  the  start: 
divide  their  pay,  $2.  A,  46  ct.;  B,  $1.06;  C,  48  ct. 

7.  A    and   B   are  partners;    A's    capital  is  $4200;   B's, 
$5600 :  after  4  months,  how  much  must  A  put  in,  to  entitle 
him  to  \  the  year's  gain  ?  $2100. 

8.  A  and   B  go   into   partnership,  each  with  $4500.     A 
draws  out  $1500,  and  B   $500,  at   the  end   of  3  mo.,  and 
each  the  same  sum  at  the  end  of  6  and  9  mo.;  at  the  end 
of  1  yr.  they  quit  with  $2200 :  how  must  they  settle? 

B  takes  $2200,  and  has  &  claim  on  A  for  $300. 

9.  A,  B,  C,  and  D  go  in  partnership ;  A  owns  12  shares 
of  the  stock;  B,  8  shares;  C,  7  shares;  D,  3  shares.     After 
3  mo.,  A  sells  2  shares  to   B,  1  to  C,  and   4  to  D;  2  mo. 
afterward,  B   sells  1  share  to  C,  and  2  to  D ;  4  mo.  after- 
ward, A  buys  2  from  C   and  2  from  D.     Divide  the  year's 
gain  ($18000). 

A,  $4650;  B,  $4650;  C,  $4700;  D,  $4000. 

10.  A,  with   $400;    B,    with   $500;   and   C,   with   $300, 
joined  in  business ;   at  the  end  of  3  mo.  A  took  out  $200 ; 
at  the  end  of  4  mo.  B  drew  out  $300,  and  after  4  mo.  more, 
he   drew  out  $150;  at  the  end  of  6  mo.  C  drew  out  $100; 
at  the  end  of  the  year  they  close ;  A's  gain  was  $225 :  what 
was  the  whole  gain  ?  $675. 


BANKRUPTCY.  331 

* 

BANKRUPTCY. 
DEFINITIONS. 

359.  1.  Bankruptcy  is  the  inability  of  a  person  or  a 
firm  to  pay  indebtedness. 

2.  A  Bankrupt  is  a  person  unable  to  pay  his  debts. 

3.  The  assets   of  a  bankrupt   are  usually  placed  in  the 
hands   of  an  Assignee,  whose  duty  it   is  to  convert  them 
into  cash,  and  divide  the  net  proceeds  among  the  creditors 
in  proportion  to  their  claims. 

REMARKS. — 1.  This  act  on  the  part  of  a  debtor  is  called  making 
an  assignment,  and  he  is  said  to  be  able  to  pay  so  much  on  the  dollar. 

2.  All  necessary  expenses,  including  assignee's  fee  (which  is 
generally  a  certain  rate  per  cent  on  the  whole  amount  of  property), 
must  be  deducted,  before  dividing. 

4.  The    amount   paid    on   a    dollar  can   be    found  by 
taking  such  a  part  of  one  dollar  as  the  whole  property  is  of 
the  whole  amount  of  the  debts;    each  creditor's  proportion 
may  be  then  found  by  multiplying  his  claim  by  the  amount 
paid  on  the  dollar. 

NOTE. — Laws  in  regard  to  bankruptcy  differ  in  the  various 
states ;  usually  a  bankrupt  who  makes  an  honest  assignment  is 
freed  by  law  of  his  remaining  indebtedness,  and  is  allowed  to  retain 
a  homestead  of  from  $500  to  $5000  in  value,  and  a  small  amount  of 
personal  property. 

EXAMPLES  FOR  PRACTICE. 

1.  A  has  a  lot  worth  $8000,  good  notes  $2500,  and  cash 
$1500;  his  debts  are  $20000:  what  can  he  pay  on  $1,  and 
what  will  A  receive,  whose  claim  is  $4500? 

SOLUTION.— $8000  +  $2500  +  $1500 —  $12000,  the  amount  of  prop- 
erty which  is  JJgflf  or  f  of  the  whole  debts.  Hence,  f  of  $1  =  GO 
ct.,  the  amount  paid  on  $1,  and  $4500  X  .60  — $2700,  the  sum  paid 
to  A. 


332  RAY'S  HIGHER  ARITHMETIC. 

2.  My  assets  are  $2520 ;  I  owe  A  $1200 ;  B,  $720 ;  C, 
$600 ;  D,  $1080 :  what  does  each  get,  and  what   is  paid  on 
each  dollar? 

A,  $840  ;  B,  $504  ;  C,  $420 ;  D,  $756;  70  ct  on  $1. 

3.  A    bankrupt's    estate     is    worth    $16000 ;    his    debts, 
$47500  ;  the  assignee  charges  5%  :  what  is  paid  on  $1,  and 
what  does  A  get,  whose  claim  is  $3650  ? 

32  ct.  on  $1,  and  $1168. 


Topical  Outline. 
PARTNERSHIP. 

1.  Definitions. 
1.  Partnership  ........       ^  j  ^ase  *•  —  Applications. 

\  Case  II.—  Applications. 
l'  Definitions. 
Applications. 


2.  Bankruptcy  ........  /  l' 

(  2. 


XYIIL  ALLIGATION. 

DEFINITIONS. 

360.  Alligation  is  the   process  of  taking  quantities  of 
different    values    in  a  combination  of    average    value.      It 
is  of  two  kinds,  Medial  and  Alternate. 

• 

ALLIGATION  MEDIAL. 

361.  Alligation    Medial    is  the  process  of  finding  the 
mean   or  average  value  of  two  or  more  things  of  different 
given  values. 

PROBLEM. — If  3  Ib.  of  sugar,  at  5  ct.  a  Ib,  and  2  lb.,  at 
4|  ct.  a  lb.,  be  mixed  with  9  lb.,  at  6  ct.  a  lb.,  what  per  lb. 

is  the  mixture  worth  ? 

OPERATION. 

SOLUTION.— The  3  lb.  at  5  ct.  Price.     Quantity.     Cost, 

a  lb.  =  15  ct,;  the  2  lb.  at  4'  ct,  5  ct.  X       3        =15  ct. 

per  lb.  =  9  ct.;  the  9  lb.  at  6  ct.  4  J     X       2        =9 

per  lb.  =  54  ct.;    therefore,  the  6       X       9         =   54 
whole  14   lb.   are   worth  78  ct.;  14  )  7  8  (  5  f  ct. 

and  78  -f- 14  —  5f  ct.  per  lb.,  Ans. 

Rule. — Find  the  values  of  the  definite  parts,  and  divide  the 
sum  of  the  values  by  the  sum  of  the  parts. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  average  price  of  6  lb.  tea,  at  80  ct.;  15  lb., 
at  50  ct.;  5  lb.,  at  60  ct.;  9  lb.,  at  40  ct.  54  ct.  per  lb. 

2.  The  average  price  of  40  hogs,  at  $8  each ;  30,  at  $10 
each  ;  16,  at  $12.50  each  ;  54,  at  $11.75  each.     $10.39  each. 

(333) 


334  RAY'S  HIGHER   ARITHMETIC. 

3.  How  fine  is  a  mixture  of  5  pwt.  of  gold,  16  carats  fine ; 
2  pwt.,  18  carats  fine;  6  pwt.,  20  carats  fine;  and  1  pwt. 
pure  gold?  18^  carats  fine. 

4.  Find  the   specific  gravity  of  a  compound  of  15  Ib.  of 
copper,  specific  gravity,  7f ;  8  Ib.  of  zinc,  specific  gravity, 
6J;  and  ^  Ib.  of  silver,  specific  gravity,  10J.  7.445— 

REMARKS. — 1.  By  the  specific  gravity  of  a  body  is  usually  under- 
stood, its  weight  compared  with  the  weight  of  an  equal  bulk  of  water;  it 
may  be  numerically  expressed  as  the  quotient  of  the  former  by  the 
latter.  Thus,  a  cubic  inch  of  silver  weighing  10J  times  as  much  as 
a  cubic  inch  of  water,  its  specific  gravity  —  lOj. 

2.  To  find  the  specific  gravity  of  a  body  heavier  than  water:  (1.) 
Find  its  weight  in  air ,  (2.)  Suspending  it  by  a  light  thread,  find  its 
weight  in  water  and  note  the  difference;  (3.)  Divide  the  first  weight 
by  this  difference.  For  example:  if  a  piece  of  metal  weighs  If  oz. 
in  air,  but  in  water  only  1J  oz.,  its  specific  gravity  =  1}  -r-  (If-  — 
1|)  =  7.  (See  Norton'*  Natural  Philosophy,  p.  152.) 

5.  What  per  cent  of  alcohol  in  a  mixture  of  9  gal.,  86% 
strong;  12  gal.,  92^  strong;  10  gal.,  95%  strong;  and  11 
gal.,  98^  strong?  93%. 

6.  At  a  teacher's  examination,  where  the  lowest  passable 
average  grade  was  50,  an  applicant   received  the  following 
grades :    In  Orthography,   50 ;  Reading,    25 ;    Writing,  50 ; 
Arithmetic,  60;  Grammar,  55;  Geography,  55:  did  he  suc- 
ceed, or  did  he  fail?  He  failed. 


ALLIGATION   ALTERNATE. 

362.  Alligation  Alternate  is  the  process  of  finding  the 
proportional  quantities  at  given   particular  prices  or  values 
in  a  required  combination  of  given  average  value. 

CASE    I. 

363.  To  proportion  the    parts,  none  of  the  quanti- 
ties being  limited . 


ALLIGATION.  335 

PROBLEM.—!.  What  relative  quantities  of  sugar,  at  9  ct. 
a  Ib.  and  5  ct.  a  lb.,  must  be  used  for  a  compound,  at  6  ct. 
alb.? 

SOLUTION. — If  you  put  1  lb.  OPERATION. 

at  9  ct.  in  the  mixture  to  be 
sold  for  6  ct.,  you  lose  3  ct.; 
if  you  put  1  lb.  at  5  ct.  in  the 


.  3  lb.  at  5  ct.  —  1  5  ct. 

.  1  lb.  at  9  ct.  =  _9  ct. 

4  lb.     worth      24  ct. 


mixture  to  be  sold  at  6  ct.,  which  is  -2^  —  6  ct.  a  lb. 

you   gain   1   ct.;    3   such   lb. 

gain  3  ct.;  the  gain  and  loss  would  then  be  equal  if  3  lb.  at  5  ct.  are 
mixed  with  1  lb.  at  9  ct. 

PROBLEM. — 2.  What  relative  numbers  of  hogs,  at  $3,  $5, 
$10  per  head,  can  be  bought  at  an  average  value  of  $7  per 
head? 

EXPLANATION. — Writing  OPERATION. 

the  average  price  7,  and  the  Diff.  Balance.  Ans. 


particular  values  3,  5,  10, 
as  in  the  margin,  we  say : 
3  sold  for  7,  is  a  gain  of  4, 


3  1 
3  1 
6  2 


which  we  write  opposite ;  5 

sold  for  7  is  a  gain  of  2 ;  10  sold  for  7  is  a  loss  of  3.  We  wish  to 
make  the  gains  and  losses  equal;  hence,  each  losing  sale  must  be 
balanced  by  one  which  gains.  To  lose  3  fours,  will  be  balanced  by 
gaining  4  threes,  and  a  gain  of  3  tu-os  will  balance  a  loss  of  2  threes. 
To  indicate  this  in  the  operation,  we  write  the  deficiency  3,  against 
the  excess  4 ;  then  the  excess  4  against  the  deficiency  3 ;  and  in 
another  column,  in  the  same  manner,  pair  the  3  and  2,  writing  each 
opposite  the  position  which  the  other  has  in  the  column  of  differ- 
ences. The  answer  might  be  given  in  two  statements  of  balance, 
thus :  for  each  3  of  the  first  kind  take  4  of  the  third,  and  for  each  3 
of  the  second  kind  take  2  of  the  third.  Since  each  balance  column 
shows  only  proportional  parts,  we  may  multiply  both  quantities  in  any 
balance  column  by  any  number,  fractional  or  integral,  and  thus  the  final 
answers  be  .  varied  indefinitely.  For  example,  had  the  second 
balancing  column  been  multiplied  by  4,  the  answer  would  have 
read,  1,  4,  4,  instead  of  1,  1,  2.  The  principle  just  stated  is  of 
great  value  in  removing  fractions  from  the  balance  columns, 
when  integral  terms  are  desired. 


336  RA  Y'S  HIGHER  ARITHMETIC. 

Rule. — 1.  Write  the  particular  values  or  prices  in  order,  in 
a  column,  having  the  smallest  at  the  head;  write  the  average 
value  in  a  middle  position  at  the  left  and  separated  from  the 
others  by  a  vertical  line. 

2.  In  another  column  to  the  right,  and  opposite  the  respective 
values,   place    in    order   the    differences   between    them    and  the 
average  value. 

3.  Then  prepare  balance  columns,  giving  to  each  of  them  two 
numbers,  one    an  excess  and  the  other  a   deficiency  taken  from 
the  difference  column;   write  each  of  these  opposite  the  position 
which  the  other  has  in  the  difference  column;    so   proceed  until 
each   number  in  the    difference    column  has  been  balanced  with 
another;  then, 

The  proportional  quantity  to  be  taken  of  each  kind,  ivill  be 
the  sum  of  the  numbers  in  a  horizontal  line  to  the  right  of  its 
excess  or  deficiency. 

NOTE.— The  proof  of  Alligation  Alternate  is  the  process  of  Alli- 
gation Medial. 

EXAMPLES  FOR  PRACTICE. 

1.  What  relative  quantities  of  tea,  worth  25,  27,  30,  32, 
and  45  ct.  per  lb.,  must  be  taken  for  a  mixture  worth  28  ct. 
per  lb.?  19,  4,  3,  1,  3  lb.  respectively. 

REMARK. — It  is  evident  that  other  results  may  by  obtained  by 
making  the  connections  differently  ;  as,  6,  17,  3,  3,  1  lb.;  or,  17,  6,  1, 
1,  3  lb. 

2.  What  of  sugar,  at  5,  5^,  6,  7,  and  8  ct.  per  lb.,  must 
be  taken  for  a  mixture  worth  6f  ct.  per  lb.? 

1,  5,  5,  7,  8  lb.  respectively;  or,  5,  1,  1,  8,  7  lb.,  etc. 

3.  What  relative  quantities  of  alcohol,  84,  86,  88,  94,  and 
96%  strong,  must  be  taken  for  a  mixture  87^  strong? 

10,  7,  3,  1,  3  gal.;  or,  7,  10,  1,  3,  1  gal.,  etc. 

4.  What  of  gold  and  silver,  whose  specific  gravities   are 


ALLIGATION. 


337 


and  10|,  will  make  a  compound  whose  specific  gravity 
shall  be  16.84?  723  Ib.  silver  to  3487  Ib.  gold. 

5.  What  of  silver  f  pure,  and  ^_.  pure,  will  make  a  mixt- 


ure 


pure? 


1  Ib.,  f  pure;  5  Ib.,  y9^  pure. 


6.  What  of  pure  gold,  and  18  carats,  and  20  carats  fine, 
must-  be  taken  to  make  22  carat  gold  ? 

1  part  18  carats,  1  part  20  carats,  3  pure. 


CASE    II. 

364.  To  proportion  the  parts,  one  or  more  of  the 
quantities,  but  not  the  amount  of  the  combination, 
being  given. 

PROBLEM. — How  many  whole  bushels  of  each  of  two 
kinds  of  wheat,  worth  respectively  $1.20  and  $1.40,  per 
bushel,  will,  with  14  bushels,  at  $1.90  per  bushel,  make  a 
combination  whose  average  value  is  $1.60  per  bushel? 


Diff.  Bal. 


OPERATION. 


Answers. 


1.60 


1.20 
1.40 
1.90 


.40 
.20 
.30 


1 

o 

3 

4 

5 

6 

7 

8 

9 

10 

19 

17 

15 

13 

11 

9 

7 

5 

3 

1 

14 

14 

14 

14 

14 

14 

14 

14 

14 

14 

SOLUTION. — We  find,  by  Case  I,  that  to  have  that  average  value 
the  parts  may,  in  one  balancing,  stand  3  of  the  first  to  4  of  the  third, 
and  in  another,  3  of  the  second  to  2  of  the  third.  By  directly  com- 
bining, we  obtain  the  proportions  1,  1,  and  2,  and  as  the  third 
must  be  14,  we  have  for  one  answer  7,  7,  14.  But  we  find  the  other 
answers  in  the  following  manner. 

The  proportion  will  not  be  altered  if  in  any  balancing  column  we 
multiply  both  quantities  by  the  same  number,  hence  the  answer  can 
be  varied  as  often  as  we  can  multiply  or  divide  the  columns,  ob- 
serving the  other  conditions,  which  are  that  the  answers  shall  be 
integral,  and  that  the  number  of  4's  and  2's  taken  shall  make  14.  As 
there  are  more  fractions  than  there  are  integers  between  any  two 
limits,  we  try  fractional  multipliers  in  order  to  obtain  the  greatest 
number  of  answers.  Observe  that  the  number  of  4's  taken  will  not 

stand  alone  in  any  answer  for  the  third  kind  of  wheat,  but  will  be 
H.  A.  20. 


338  It  A  Y'S  HIGHER  ARITHMETIC. 

added  to  some  number  of  2's;  the  number  of  3's  taken  as  any 
one  answer  for  the  first  or  second  kind  will  not  be  increased  by  any 
other  product ;  hence,  if  we  use  a  fractional  multiplier,  it  must  be 
such  that  its  denominator  will  disappear  in  multiplying  by  3 ;  and 
this  shows  that  a  fractional  multiplier  will  not  be  convenient  unless 
it  can  be  expressed  as  thirds.  Therefore,  the  remaining  question  is, 
How  many  thirds  of  4  with  thirds  of  2  will  make  14?  Since  14  = 
-432,  the  question  is  the  same  as  to  ask,  How  many  whole  4's  with  whole 
2's  will  make  42  ?  It  is  plain  that  there  can  not  be  more  than  ten 
4's.  We  can  take  J  of 

1  four  and  19  twos,  or  J  of  first  column  and  -^  of  second. 

2  fours   "     17       "       "    |   "      "          "  "    -V    "        « 

o        tt         it      15        it        it     3     tt       it  tt  it      ij>_    a  u 

The  answers  are  now  obvious  :  write  1,  2,  3,  etc.,  parallel  with  19, 
17,  15,  etc.,  and  the  14's  in  the  third  row. 

Rule. — Find  the  proportional  parts,  as  in  Case  I,  and  ob- 
serve the  term  or  terms  in  the  balance  columns,  standing  oppo- 
site the  value  or  price  corresponding  to  the  limited  quantity; 
then  find  what  multipliers  will  produce  the  given  limited  quan- 
tity in  the  required  place,  and  of  those  multipliers  use  only 
such  as  will  agree  with  the  remaining  conditions  of  the  problem. 


EXAMPLES  FOR  PRACTICE. 

'  1.  How  many  railroad  shares,  at  50%,  must  A  buy,  who 
has  80  shares  that  cost  him  72%,  in  order  to  reduce  his 
average  to  60%?  96  share? 

2.  How  many  bushels  of  hops,  worth  respectively  50,  60 
and  75  ct.  per  bu.,  with  100  bu.,  at  40  ct.  per  bu.,  will  mak 
a  mixture  worth  65  ct.  a  bu.?  2,  2,  and  254 

3.  How  much  water  (0  per  cent)  will  dilute  3  gal.  2  qt. 
1  pt.  of  acid  91%  strong,  to  56^  ?  2  gal.  1  qt.  \  pt. 

4.  A  jeweller  has  3  pwt.  9  gr.  of  old  gold,  16  carats  fine : 
how  much  U.  S.  gold,  2 If  carats  fine,  must  he  mix  with  it, 
to  make  it  18  carats  fine?  1  pwt.  21  gr. 


ALLIGATION.  339 

5.  How  much  water,  with  3  pt.  of  alcohol,   96%   strong, 
and  8  pt.,  78^,  will  make  a  mixture  60%  strong?       4^-  pt. 

6.  I  mixed  1  gal.  2  qt.  ^  pt.  of  water  with  3  qt.  1£  pt.  of 
pure  acid;  the  mixture  has  15%    more  acid    than   desired: 
how  much  water  will  reduce  it  to  the  required  strength? 

1  gal.  2  qt.  11  pt. 

7.  How  much  lead,  specific  gravity  11,  with  ^  oz.  copper, 
sp.  gr.  9,  can  be  put  on  12  oz.  of  cork.  sp.  gr.  ^,  so  that  the 
three  will  just  float,  that  is,  have  a  sp.  gr.  (1)  the  same  as 
water  ?  2  Ib.  7|  oz. 

8.  How  many  shares  of  stock,  at  40%,  must  A  buy,  who 
has  bought  120  shares,  at  74^,  150  shares,  at  68^,  and 
130  shares,  at  54%,  so  that  he  may  sell  the  whole  at  60^, 
and  gain  20%  ?  610  shares. 

9.  A  buys  400  bbl.   of  flour,  at  $7.50  each,  640  bbl.,  at 
$7.25,  and  960  bbl.,  at  $6.75:  how  many  must  he  buy  at 
$5.50,  to  reduce  his  average  to  $6.50  per  bbl.?         1120  bbl. 

CASE   III. 

365.  To  proportion  the  parts,  the  amount  of  the 
whole  combination  being  given. 

PROBLEM. — If  a  man  pay  $16  for  each  cow,  $3  for  each 
hog,  and  $2  for  each  sheep,  how  many  of  each  kind  may  he 
purchase  so  as  to  have  100  animals  for  $600  ? 

SOLUTION.— Pro-  OPERATION. 

ceeding  as  in  Case  Diff.       Bal.  Answers. 

I,  we  find  that  the 


given    average    re-  $  6 

quires  5  of  the  first 


2 

3 

16 


4 
3 
10 


12 
64 
24 


25 
50 
25 


38 
36 
26 


51 
22 

27 


64 


28 


kind  with  2  of  the  713 

third,  and  10  of  the 

second  kind  with  3  of  the  third  ;  taken  in  two  parts,  7  are  required 

in  one  balancing,  and  13  in  another;  these  being  together  20,  which 

is  contained  five  times  in  the  required  number,  100,  if  we  multiply  all 

of  the  terms  in  the  balance  columns  by  5,  we  have  for  one  answer 

25  sheep,  50  hogs,  and  25  cows. 


340  RAY'S  HIGHER  ARITHMETIC. 

As  there  are  other  multipliers  affording  results  within  the  con- 
ditions, we  leave  the  student  to  find  the  remaining  answers  by  a 
process  similar  to  that  shown  under  Case  II. 

REMARK. — Suppose  that  we  had  to  determine  400 

how   many  7's  and  ll's  would   make  400.     By  2       22 

trial,  we  find  that  two  ll's  taken  away  leave  an  3  7  8.. .5  4 

exact  number  of  7's.     It  is  now  unnecessary  to  77 

take  single  ll's  or  proceed  by  trial  any  farther ;  9    3  01... 4  3 

for,  as  378  is  an  exact  number  of  7's,  if  we  take  7  7 

away  ll's  and  leave  7's,  we  must  take  seven  ll's ;  ^  g    2  24.. .3 2 
and  thus  the  law  of  continuation  is  obvious  :  400 
is  composed  of 

ll's 2,      9,      16,      23,      30; 

With    7's. .....54,    43,      32,      21,      10. 

Rule. — Proportion  the  parts  as  in  Case  I;  then,  noting  the 
sums  of  the  balancing  columns,  find,  by  trial  or  by  direct  di- 
vision, what  multipliers  ivill  make  those  columns  together  equal 
to  the  given  amount  of  the  combination,  and  of  those  multipliers 
use  only  such  as  will  agree  ivith  the  remaining  conditions. 


EXAMPLES  FOR  PRACTICE. 

1.  What  quantities  of  sugar,  at  3  ct.    per  Ib.  and  7  ct. 
per  Ib.,  with  2  Ib.  at  8  ct.,  and  5  Ib.  at  4  ct.  per  Ib.,  will 
make  16  Ib.,  wrorth  6  ct.  per  Ib.? 

f  Ib.  at  3  ct.,  8J  Ib.  at  7  ct. 

2.  How  many  bbl.  flour,  at  $8  and  $8.50,  with  300  bbl. 
at  $7.50,  and  800  at  $7.80,  and  400  at  $7.65,  will  make 
2000  bbl.  at  $7.85  a  bbl.? 

200  bbl.,  at  $8 ;  300  bbl.,  at  $8.50 

3.  What  quantities  of  tea,  at  25  ct.  and  35  ct.  a  Ib.,  with 
14  Ib.  at  30  ct.,  and  20  Ib.  at   50  ct.,  and  6  Ib.  at  60  ct., 
will  make  56  Ib.  at  40  ct.  a  Ib.? 

10  Ib.  at  25  ct.,  and  6  Ib.  at  35  ct. 


TOPICAL   OUTLINE.  341 

4.  Hex;p   much  copper,   specific   gravity   7|,    with   silver, 
specific  gravity  10^,  will  make  1  Ib.  troy,  of  specific  gravity 
8f  ?  7fff  oz.  copper,  4||§  oz.  silver. 

5.  How  much  gold  15  carats  fine,  20   carats   fine,   and 
pure,   will    make   a   ring   18   carats    fine,  weighing   4  pwt. 
16  gr.?  2  pwt.  16  gr.;  1  pwt.;   1  pwt. 

6.  A  dealer  in  stock  can   buy  100  animals  for  $400,  at 
the  following  rates, — calves,  $9;   hogs,  $2;  lambs,  $1:  how 
many  may  he  take  of  each  kind  ? 

37  calves,  4  hogs,   59  lambs; 
(one  of  nine  different  answers.) 

7.  Hiero's  crown,  sp.  gr.  14f ,  was   of  gold,  sp.  gr.   19^, 
and  silver,  sp.  gr.  10^;  it  weighed  17 J  Ib.:  how  much  gold 
was  in  it?  lOff  Ib. 


Topical  Outline. 
ALLIGATION. 

f  1.  Definitions. 
1.  Alligation I  C  L  Alligatiou 


Medial /  L  Definitions. 


2-  Killds"  ^  i   2.  Rules. 


2.  Alligation 


Alternate..  „ 

2.  Cases.. 


1.  Definitions. 


fl.  Rule. 
J  II.  Rule. 
I  III.  Rule. 


XIX.   IN  VOLUTION. 

DEFINITIONS. 

366.  1.  A  power  of  a  quantity  is  either  that  quantity 
itself,  or  the  product  of  a  number  of  factors  each  equal  to 
that  quantity. 

KEMARK. — Kegarding  unity  as  a  base,  we  may  say,  the  power  of 
a  quantity  is  the  product  arising  from  taking  unity  once  as  a  mul- 
tiplicand, with  only  the  given  quantity  a  certain  number  of  times 
as  a  factor.  The  power  takes  its  name  from  the  number  of  times 
the  quantity  is  used  as  factor.  Unity  is  no  power  of  any  other 
positive  number. 

2.  The  root  of  a  power  is  one  of  the  equal  factors  which 
produce  the  power. 

3.  Powers  are  of  different  degrees,  named  from  the  number 
of  times  the  root  is  taken  to  produce  the  power.     The  degree 
is  indicated  by  a  number  written  to  the  right  of  the   root, 
and  a  little  above;  this  index  number  is  called  an  exponent. 
Thus, 

5  X  5,          or  the  2d  power  of  5,  is  written  52. 
5X5X5,      "       3d     "       "      "        "      53. 

4.  The  second  power  of  any  number  is  called  the  square, 
because  the  area  of  a  square  is  numerically  obtained  by  form- 
ing a  second  power. 

5.  In  like  manner  the  third  power  of  any  number  is  called 
its  cube,  because  the  solidity  of  a  cube  is  numerically  ob- 
tained by  forming  a  third  power. 

367.  To  find  any  power  of  a  number,  higher  than 
the  first, 

(342) 


INVOLUTION.  343 

Rule. — Multiply  the  number  by  itself,  and  continue  the  mul- 
tiplication till  that  number  has  been  used  as  factor  as  many 
times  as  are  indicated  by  the  exponent. 

NOTES. — 1.  The  number  of  multiplications  will  be  one  less  than 
the  exponent,  because  the  root  is  used  twice  in  the  first  multiplica- 
tion, once  as  multiplicand  and  once  as  multiplier. 

2.  When  the  power  to  be  obtained  is  of  a  high  degree,  multiply 
by  some  of  the  powers  instead  of  by  the  root  continually ;  thus,  to 
obtain  the  9th  power  of  2,  multiply  its  6th  power  (64)  by  its  3d 
power  (8);  or,  its  5th  power  (32)  by  its  4th   power  (16):  the  rule 
being,  that  the  product  of  any  two  or  more  powers  of  a  number  is  thai 
power  whose  degree  is  equal  to  the  sum  of  their  degrees. 

3.  Any  power  of  1  is  1 ;  any  power  of  a  number  greater  than  1  is 
greater  than  the  number  itself :  any  power  of  a  number  less  than  1,  is 
less  than  the  number  itself. 

368.  From  Note  2,  43  X  43  X  43  X  43  X  43  =  41'5;  but 
the  expression  on  the  left  is  the  5th  power  of  43;   hence, 
(43)5  =415;  that  is,  when  the  exponent  of  the  power  required 
is  a  composite  number    (15),  raise  the  root  to   a  power  whose 
exponent  is  one  of  its  factors  (3),   and  this  result  to  a  power 
whose  exponent  is  the  other  factor  (5). 

NOTE. — Let  the  student  carefully  note  the  difference  between 
raising  a  power  to  a  power 9  and  multiplying  together  different  powers  of 
the  same  root ;  thus, 

23  X22  =25. 

Here  we  have  multiplied  the  cube  by  the  square  and  obtained  the 
5th  power ;  but  the  5th  power  is  not  the  square  of  the  cube;  this  is  the 
sixth  power,  and  we  write 

(22)3  =  26,  or  (23)2  =  23X2  =26. 

369.  Any  power  of  a  fraction  is  equal  to  that  power  of 
the  numerator  divided  by  that  power  of  the  denominator. 

370.  The  square  of  a  decimal  must  contain  twice,  and  its 
cube,  three   times  as  many  decimal  places  as  the  root,  etc.; 


344 


It  A  Y'S  HIGHER  ARITHMETIC. 


hence,  to  obtain  any  power  of  a  decimal,  we  have  the  follow- 
ing rule: 

Rule. — Proceed  as    if  the   decimal  were  a  whole   number, 

and  point  off  in  the    result   a  number  of  decimal  places  equal 

to  the  number  in   the   root  multiplied  by  the  exponent  of  the 
power. 

EXAMPLES  FOR  PRACTICE. 

Show  by  involution,  that: 

1.  (5)2 

2.  143 

3.  65 

4.  1922 

5.  I10 

6.  (!)4 

7. 


processes    for    squaring    and    cubing 


equals   25. 

8.  (|)5 

equals   if-flrf- 

2744. 

9.  (.02)s 

.000008 

7776. 

10.  (54)2 

390625. 

"   36864. 

11.  (.046)s 

"  .000097336 

1. 

12.  (i)7 

4782969- 

AV 

13.  20562 

"   4227136. 

"   Hfl- 

14.  (7.621) 

2  "   58.1406}- 

371.     Special 
numbers. 


PROBLEM.— Find  the  square  of  64. 


PARALLEL  OPERATIONS. 

tens  +  4  units. 


64  = 
64 

60         +4  =  6  t 
60         +4 

256  = 
3840=    602- 

60X4  +  42 
[-60X4 

4096=    602 +  2(60X4)  +  42 
=  3600  +  480  +  16. 

The  operations  illustrate  the  following  principle : 

PRINCIPLE.  —  The  square  of  the  sum  of  two  numbers  is  equal 
to  the  square  of  the  first,  plus  twice  the  product  of  the  first  by 
the  second,  plus  the  square  of  the  second.  Thus : 


252  =  (22  +  3)  2  =  484  +  2  (22  X  3) 

252  =  (21  +  4)2  =441  +  2  (21  X  4)  +  16  =  625. 

252  =  (20  +  5)2  =-400  +  2  (20  X  5)  +  25  =  625. 

REMARK.—  The  usual  application  of  this  principle  in  Arithmetic 
is,  in  squaring  a  number  as  composed  of  tens  and  units.  The  third 
statement  above  illustrates  this;  and,  if  we  represent  the  tens  by  t, 
the  units  by  w,  we  have  the  following  statement  : 

(t  +  u)2  =  t2  +  2tu  +  w2  ;  or,  in  common  language  : 

The  square  of  any  number  composed  of  tens  and  units  is  equal  to  the 
square  of  the  tens,  +  twice  the  product  of  the  tens  by  the  units,  +  the  square 
of  the  units. 


EXAMPLES  FOR  PRACTICE. 

Square  the  following  numbers,  considering  each  as  the 
sum  of  two  quantities,  and  applying  the  principle  announced 
in  Art.  371,  on  page  preceding: 


1.  19.  361. 

2.  29.  841. 

3.  4.  16. 


4.  40.  1600. 

5.  125.  15625. 

6.  59.  3481. 


ILLUSTRATION. — Draw  a  square.  From  points  in  the  sides,  at 
equal  distances  from  one  of  the  corners,  draw  two  straight  lines 
across  the  figure,  each  parallel  to  two  sides  of  the  figure.  These 
two  lines  will  divide  the  square  into  four  parts,  two  of  them  being 
squares  and  two  of  them  rectangles.  The  base  being  composed  of 
two  lines,  and  the  square  of  four  parts,  we  see  that 

372.  The  square  described  on  the  sum  of  two  lines  is 
equal  to  the  sum  of  the  squares  described  on  the  lines,  plus 
twice  the  rectangle  of  the  lines. 

REMARK. — Both  the  principle  employed  above  and  the  illustra- 
tion are  frequently  used  in  explaining  the  method  for  square  root. 


346  RAY'S  HIGHER  ARITHMETIC. 

PROBLEM.  —  Find  the  cube  of  64. 

PARALLEL   OPERATIONS. 

642  =       4096=  602  +  2(60    X4)  +  4 

_  64=  _  60  +  4 


16384=  602X4+2(60X42) 

24576    =60 


The  operation  illustrates  the  following  principle  : 

PRINCIPLE.  —  The  cube  of  any  number  composed  of  two  parts, 
is  equal  to  the  cube  of  the  first  part,  plus  three  times  the  square 
of  the  first  by  the  second,  plus  three  times  the  first  by  the  square 
of  the  second,  plus  Hie  cube  of  the  second.  Thus  : 

253  =  (22  +  3)3  =  22s  +  3  (222  X  3)  +  3  (22  X  32)  +  33 
=  10648  -f  4356  +  594  +  27  =  15625. 

REMARK.  —  The  usual  application  of  this  principle  in  Arithmetic 
is,  in  the  cubing  of  a  number  as  composed  of  tens  and  units.  Repre- 
senting the  tens  by  t,  and  the  units  by  u,  we  have  the  following  state- 
ment, which  the  student  will  express  in  common  language,  similar 
to  that  of  the  principle  used  in  squaring  numbers  : 

(t  +  uy  =  t3 


EXAMPLES  FOR  PRACTICE. 

Considering  the  following  numbers  as  made,  each,  of  two 
parts,  cube  them  by  the  principle  just  stated : 


(1.)  19.  6859. 
(2.)  29.  24389. 
(3.)  4.  64. 


(4.)  40.  64000. 
(5.)  125.  1953125. 
(6.)  216.  10077696. 


XX.  EVOLUTION. 

DEFINITIONS. 

373.  1.  Evolution  is  the  process  of  finding  roots  of 
numbers. 

2.  A  root  of  a  number  is  either  the  number  itself  or  one 
of  the  equal  factors  which,  without  any  other  factor,  produce 
the  number. 

Since  a  number  is  the  first  root,  as  also  the  first  pouer  of  itself, 
no  operation  is  necessary  to  find  either  of  these ;  hence,  in  evolu- 
tion, we  seek  only  one  of  the  equal  factors  which  produce  a  power. 

Evolution  is  the  reverse  of  Involution,  and  is  sometimes  called 
the  Extraction  of  Roots. 

3.  Roots,  like  powers,  are  of  different  degrees,  2d,  3d,  4th, 
etc.;  the  degree  of  a  root  is  always  the  same  as  the  degree 
of  the  power  to  which  that  root  must  be  raised  to  produce 
the  given  number. 

Thus,  the  3d  root  of  343  is  7,  since  7  must  be  raised  to  the  3d 
power,  to  produce  343 ;  the  5th  root  of  1024  is  4,  since  4  must  be 
raised  to  the  5th  power,  to  produce  1024. 

Since  the  2d  and  3d  powers  are  called  the  square  and  cube,  so  the 
2d  and  3d  roots  are  called  the  square  root  and  cube  root. 

4.  To  indicate  the  root  of  a  number,  we  use  the  Radical 
Sign  (]/),  or  a  fractional  exponent. 

The  radical  sign  is  placed  before  the  number;  the  degree 
of  the  root  is  shown  by  the  small  figure  between  the  branches 
of  the  radical  sign,  called  the  Index  of  the  root. 

Thus,  ^18  signifies  the  cube  root  of  18 ;  jX9  signifies  the  5th 
root  of  9.  The  square  root  is  usually  indicated  without  the  index 

2 ;  thus,  1/10  is  the  same  as  £"10. 

(347) 


348  RAY'S  HIGHEfi  ARITHMETIC. 

5.  The  root  of  a  number  may  be  indicated  by  a  fractional 
exponent  whose  numerator  is   1,  and   ivhose   denominator  is  the 
index  of  the  root  to  be  expressed. 

Thus,  1/7  =  72,  and  T3//5  =  51 ;  similarly,  4f  =  16T,  the  numera- 
tor indicating  a  power  and  the  denominator  a  root. 

» 

6.  A   perfect    power    is  a  number   whose  root  can   be 
exactly  expressed  in   the   ordinary  notation;    as  32,   whose 
fifth  root  is  2. 

7.  An  imperfect  power  is  a  number  whose  root  can  not 
be  exactly  expressed  in  the  ordinary  notation;  as  10,  whose 
square  root  is  3. 1622 -f- 

8.  The  squares  and  -cubes  of  the  first  nine  numbers  are 
as  follows: 

Numbers,  123456789 
Squares,  1  4  9  16  25  36  49  64  81 
Cubes,  1  8  27  64  125  216  343  512  729 

9.  The  Square  Root  of  a  number  is  one  of  the  two  equal 
factors  which,  without  any  other  factor,  produce  that  num- 
ber ;  thus,  7  X  7  =  49,  and  1/49  =  7. 

10.  The   Cube  Root  of  a   number  is  one  of  the  three 
equal  factors  which,  without  any  other,  produce  that   num- 
ber; thus,  3  X  3  X  3  =  27,  and  1^27  =  3. 

374.  Concerning  powers  and  roots  in  the  ordinary  deci- 
mal notation,  we  state  the  following  principles : 

PRINCIPLES. — 1.  The  square  of  any  number  has  tivice  as 
many,  or  one  less  than  twice  as  many,  figures  as  the  number 
itself  has. 

2.  There  will  be  as  many  figures  in  the  square  root  of  a 
perfect  power  as  there  are  periods  of  two  figures  each  in  the 
power,  beginning  with  units,  and  also  a  figure  in  the  root  cor- 
responding to  a  part  of  such  period  at  the  left  in  the  power. 


EVOLUTION.  349 

3.  The  cube  of  a  number  has  three  times  as  many  figures, 
or  one  or  tivo   less    than    'three   times  as   many,  as  the  number 
itself  has. 

4.  There  will  be  as  many  figures  in  the  cube  root  of  a  perfect 
power,  as  there  are  periods  of  three  figures  each  in  the  power, 
beginning  with  units,  and  also  a  figure  corresponding  to  any  part 
of  such  a  period  at  the  left  hand. 


EXERCISES. 

1.  Prove  that  there  will  be  six  figures  in  the  cube  of  the 
greatest  integer  of  two  figures. 

2.  Prove  that  there  will    be  twelve  figures  in  the  fourth 
power  of  the  greatest  integer  of  three  figures. 


EXTRACTION  OF  THE  SQUARE  ROOT. 
FIRST  EXPLANATION. 

PROBLEM. — What  is  the  length  of  the  side  of  a  square 
containing  576  sq.  in.? 

SOLUTION. — The  length  required   will  OPERATION. 

be  expressed  by  the  square  root  of  576;  576(20 

by  Principle  2,  we  know  that  the  root  can  4 


400    2 4  in.,  Ans. 

176 

176 


have  no  less  than  two  places  of  figures;        40 
and  since  the  square  of  3  tens  is  greater,  4 

and  that  of  2  tens  less  than  576,  the  root  |~^ 
must  be  less  than  30  and  greater  than  20; 
hence,  2  is  the  first  figure  of  the  root,  and  400  the  greatest  square  of 
tens  contained  in  576.  Let  the  first  of  the  accompanying  figures 
represent  the  square  whose  side  is  to  be  found.  We  see  that  the 
side  must  be  greater  than  20,  and  that  the  given  area  exceeds  by 
176  sq.  in.  the  square  whose  side  is  20  in.  long.  It  is  also  evident 
from  the  figure  that  the  176  sq.  in.  may  be  regarded  as  made  of  three 
parts,  two  of  them  being  rectangles  and  one  a  small  square ;  these 
parts  are  of  the  same  width,  and,  if  that  width  be  ascertained  and 


350 


RAY'S  HIGHER  ARITHMETIC. 


20X4  =  80 


20X20=400 


added  to  20  in.?  the  required  side  will  be  found.     The  two  20-inch 

rectangles,  with  the  small  square,  may  be  considered  as  making  one 

long  rectangle  of  the  required 

width,  as  shown  in  the  figure 

on  the  right ;  and,  as  the  exact 

area  of  that  rectangle  is  176 

sq.  in.,  if  we  knew  its  length, 

its  true  width  would  be  found 

by    dividing   the    area    by   the 

length   (Art.  197,  7);  but  we 

do    know    that    the  length    is 

greater  than  40  in.,  and  hence, 

that  the   width    is,  in   inches, 

less  than  the  quotient  of  176  by  40;  and  since,  in  176,  40  is 

contained  more  than  four  times,  but  not  jive  times,  4  is  the 

highest  number  we  need  try  for  the  width.     Now,  as  the  true 

length  of  that  rectangle  is  40  in.  increased  by  the  true  width,  the 

proper  way  to  try  4,  is  to  add  it  to  40  and  multiply  the  sum 

by  4;  thus,  40  in.+  4  in.  =  44  in.;  and  44  X  4  —  176.     This 

shows  that  4  in.  is  the  width  of  the  rectangle,  and  hence  the  required 

side  is  20  in.  -f-  4  in.  =  24  in.,  Ans. 

REMARKS. — 1.  Since  17  contains  4  as  often  integrallyj  as  176  con- 
tains 40,  it  is  convenient  to  use  simply  the  17  as  dividend  with  4  as 
divisor,  and  then  annex  the  quotient  to  the  divisor  and  to  the  first 
figure  of  the  root. 

2.  At  the  first  step  we  ascertained  that  the  whole  root  was  greater 
than  2  tens  and  less  than  3  tens ;  at  the  next  step  we  learned  that 
the  units  were  not  equal  to  5,  and  by  trial  they  were  found  to  be  4. 
The  whole  process  was  a  gradual  approach  to  the  exact  root, — one 
figure  at  a  time.     It  is  important  for  the  student  to  note  that  in  the 
processes  of  evolution  there  must  be  steps  of  trial.     Even  the  higher 
branches  will  not  exempt  from  all  trial  work.     The  most  valuable 
rules  pertaining  to  such  numerical  opera- 
tions, simply  narrow  the   trial  by  making 

the  limits  obvious.  Thus  our  device  above 
showed  the  second  part  of  the  root  less 
than  5;  an  actual  trial  showed  it  to  be 
exactly  4. 

3.  If    the    power    had    been    58081,  we 
should  have  found  there  were  three  figures 

in  the  root ;  here,  as  in  the  former  case,  4  is  found  the  greatest  figure 


44 


58081(241 

4_ 

180 

176 


481 


481 
481 


EVOLUTION.  351 

which  can  stand  in  ten's  place,  and  we  may  treat  the  24  tens  exactly 
as  we  treated  the  2  tens  in  the  first  illustration. 


SECOND  EXPLANATION. 

We  learned  in  Art.  371  that  the  square  of  a  number  composed  of 
tens  and  units  is  equal  to  the  square  of  the  tens,  plus  twice  the 
product  of  the  tens  by  the  units,  plus  the  square  of  the  units. 

The  square  of  (20  +  4),  or  242,  is  202  +  2  X  (20  X  4)  +  42. 
Now,  if  the  square  of  the  tens  be  taken  away,  there  will  remain  2  X 
(20  X  4)  +  42  =  40  times  4,  and  4  times  4,  or,  simply  (40  +  4)  X  4. 
We  see  then  that  if  the  square  of  the  tens  be  taken  away,  the  remain- 
der is  a  product  whose  larger  factor  is  the  double  of  the  tens,  increased 
by  the  units,  the  smaller  factor  being  simply  the  units. 

Suppose,  then,  in  seeking  the  square  root  of  1764,  we  have  found 
the  tens  of  the  root  to  be  4;  the  remainder  164  must  be  the  product 
of  the  units  by  a  factor  which  is  equal  to 
the  sum  of  twice  the  tens  and  once  the  OPERATION. 

units.     If  we  knew  the  units,  that  larger  1764(40 

factor  could  be  found  by  doubling  the  tens  1600 

and   adding    the    units;    if,  on  the   other          80 
hand,  we  knew  the  larger  factor,  the  units  2 

could  be  found  by  direct  division ;  we  do         #  2 
know  that  larger  factor  to  be  more  than  80, 

and  hence  that  the  units  factor  is  less  than  the  exact  number  of  times 
164  contains  80.  Therefore,  the  units  figure  can  not  be  so  great  as 
3,  and  the  largest  we  need  try  is  2.  The  proper  way  to  try  2,  is  to 
add  it  to  80,  and  then  multiply  by  2;  this  being  done,  we  see  that, 
the  product  being  equal  to  164,  2  is  the  exact  number  of  units,  80  + 
2  the  larger  factor  exactly,  and  42  the  exact  root. 


164    42,  An*. 
\  64 


NOTE. — These  successive  steps  showed,  that  the  first  figure  was 
the  root  as  nearly  as  tens  could  express  it ;  with  the  second  figure 
we  found  the  root  exactly.  Had  the  power  been  1781,  the  42 
would  still  have  been  the  true  root  as  far  as  tens  and  units  could 
express  it;  and  at  the  next  step,  seeking  a  figure  in  tenth's  place, 
we  would  have  found  the  true  root,  42.2,  as  far  as  expressible  by 
tens,  units,  and  tenths.  Continuing  this  operation,  we  find  42.201895 
to  be  the  root,  true  as  far  as  millionths  can  express  it;  so,  in  any  case, 
when  a  figure  is  correctly  found,  the  true  root  can  not  differ  from  the 
whole  root  obtained,  by  so  much  as  a  unit  in  the  place  of  that  jigure. 


352  RAY'S  HIGHER  ARITHMETIC. 

375.  To  extract  the  square  root  of  a  number  writ- 
ten in  the  decimal  notation,  as  integer,  fraction,  or 
mixed  number.  * 

Rule. — 1.  Point  off  the  number  into  periods  tf  two  figures 
each,  commencing  with  units. 

2.  Find  the  greatest  square  in  the  first  period  on  the  left; 
place  its  root  on  the  right,  like  a  quotient  in  division;  subtract 
the  square  from  the  period,   and  to  the  remainder  bring  down 
the  next  period  for  a  dividend. 

3.  Double  the  root  already  found,  as  if  it  were  units,  and 
write  it  on  the  left  for  a  trial  divisor ;  find  how  often  this  is 
contained  in  the  dividend,   exclusive  of  the  right-hand  figure, 
annexing  the  quotient  to  the  root  and  to  the  divisor;   then  mul- 
tiply the  complete  divisor  by  the  quotient,  subtract  the  product 
from  the  dividend,  and  to  the  remainder  bring  down  the  next 
period  as  before. 

4.  Double  the  root  as  before,  place  it  on  the  left  as  a  trial 
divisor,  proceeding  as   with    the  former    divisor    and  quotient 
figure;   continue  the  operation  until  the  remainder  is  nothing, 
or  until  the  lowest  required  decimal  order  of  the  root  has  been 
obtained. 

NOTES. — 1.  If.  any  product  be  found  too  large,  the  last  figure  of  the 
root  is  too  large. 

2.  The   number  of  decimal  places   in   the  power  must  be  even; 
hence    the  number    of  decimal    periods   can  be   increased   only  by 
annexing  ciphers  in  pairs.     Contracted  division  may  be  used  to  find 
the  lower  orders  of  an  imperfect  root. 

3.  When  a  remainder  is  greater  than  the  previous  divisor  it  does 
not  follow  that  the  last  figure  of  the  root  is  too  small,  unless  that 
remainder  be  large  enough  to  contain  twice  the  part  of  the  root  already 
found   and   1   more;    for   this   would  be   the   complete  divisor,    and 
would  be  contained  in  the  remainder  if  the  root  were  increased  by  1. 
Hence,  the  square  of  any  number  must  be  increased  by  a  unit  more  than 
twice  the  number  itself,  to  make  the  square  of  the  next  higher. 

Thus,  1252  =15625;   simply  add  250+1,  and  find   15876=1262. 


EVOLUTION. 


353 


EXAMPLES  FOR  PRACTICE. 


1856. 

240. 

4062. 

7007. 
270.194 
7583.69 


13.  Find   the  square  root  of  3,  true  to  the  7th  decimal 
place.  1.7320508 

14.  Find  the  square  root  of  9.869604401089358,  true  to 
the  7th  decimal  place.  3.1415926 

15.  i/,030625  X  1/4O96"X  1/.00000625  —  what?    .0028 

16.  ]/(126)2  X  (58)2  X 


1.  i/2809. 

53. 
38. 
109. 
13625. 
8944.9 
2490.74 

7.  1/3444736. 

2.  1/1444. 

8.  1/57600. 

3.  1/11881. 

9.  1/16499844. 

4.  i/185640625. 

10.  1/49098049. 

5.  1/80012304. 

11.  1/73005. 

6.  1/6203794. 

12.  1/3863. 

17.  1/12.96  X  sq.  rt.  of  ^  =  3.2863 


KEMARK.  —  The  remainder,  at  any  point,  is  equal  to  the  square  of 
an  unknown  part,  plus  twice  the  product  of  that  part  by  a  known  part. 
The  remainder  may  also  be  considered  as  the  difference  of  two  squares, 
which  is  always  equal  to  the  product  of  the  sum  of  the  roots  by  their 
difference. 

376.  The  square  root  of  the  product  of  any  number  of 
quantities    is  equal  to    the  product  of  their  square   roots; 
thus,  1/16  X  .49  =  4  X  .7  =  2.8 

377.  The  square  root  of  a  common  fraction  is  equal  to 
the  square   root   of  the  numerator,  divided    by  the  square 
root  of  the  denominator. 

REMARK.  —  It  is  advantageous  to  multiply  both  terms  by  what 
will  render  the  denominator  a  square. 


EXAMPLES  FOR  PRACTICE. 


1.  y-f-      =  .92582+ 

2.  i/34f*=  5.8843+ 

H.  A.  30. 


—  f  nearly. 


3. 


4.  i/272&"=  161 


354 


RAY'S  HIGHER  ARITHMETIC. 


5.  i/6J    =  2.5298+ 

6.  T/tf  X  A¥7  Xlf  X  T/T=  .45886+ 

7.  T/123.454321  X  .81  =  9.9999 

8.  i/1.728x4.8Xfr  =  - 
1/2L 


X  1/21,  written  also 


EXTRACTION  OF  THE  CUBE  KOOT. 
FIRST  EXPLANATION. 

PROBLEM.  —  What  is  the  edge  of  a  cube  whose  solid  inches 
number  13824? 

OPERATION. 

13824(20 
8000   4 


202X3      =1200 
20 


=    16 


1456 


5824  24,  Ans. 


5824 


SOLUTION. — The  length 
required  will  be  expressed 
by  the  cube  root  of  13824. 
By  Prin.  4,  we  know  the 
root  can  have  no  less  than 
two  places  of  figures;  also, 
since  the  cube  of  3  tens  is 
greater  and  that  of  2  tens  is  less  than  13824,  the  root  must  be 
less  than  30  and  more  than  20 ;  consequently,  2  is  the  first  figure  of 
the  root,  and  8000  is  the  greatest  cube  of  tens  contained  in  13824.  Let 
the  first  of  the  accompanying  figures 
represent  the  cube  whose  edge  is  to 
be  found.  We  see  that  it  must  be 
greater  than  20  inches,  and  that  the 
given  solidity  exceeds  by  5824  cu.  in. 
the  cube  whose  edge  is  20  inches,  and 
which  for  convenience  we  will  call  A. 
(See  Fig.  2. )  It  is  also  evident  from  the 
second  figure,  where  the  separate  parts 
are  shown,  that  the  5824  cu.  in.  may  be 
regarded  as  made  up  of  seven  parts, 
three  of  them  being  square  blocks 

(  B)  20  in.  long,  three  of  them  being  rectangular  blocks  (  0 )  of  the 
same  length,  and  one  a  small  cube  (  C).  These  parts  are  of  the 
same  thickness,  and  if  that  thickness  be  ascertained  and  added  to  20  in., 
the  required  edge  will  be  found.  The  square  blocks  and  the  oblong 
blocks,  with  the  small  cube,  may  be  considered  as  standing  in  line 


Fig.  1. 


EVOLUTION. 


355 


(Fig.  3)  and  forming  one  oblong  solid  of  uniform  thickness.     Now,  as 

the  exact  solidity  of  that  solid 

is  5824  cu.  in.,  if  we  knew  its 

side  surface,  its  true  thickness 

would  be  found  by  dividing  the 

number  expressing  the  solidity 

by  the  number  expressing  the 

surface.    But  we  do  know  that 

side  surface  to  be  greater  than 

3  times  400  sq.  in.,  and  hence 
the  thickness  must  be  less  than 
the  quotient  of  5824  by  1200; 
and  since  in  5824,  1200  is  con- 
tained 4  times  but  not  5  times, 

4  is  the  highest  number  we  need 
try  for  the  width  ;  as  the  exact 
surface  of  one  side  is  equal  to 
1200   sq.   in.,    increased   by   3 
rectangles  20  in.  long,  and  a 
small  square  also,  each   of   a 
width   equal   to    the  required 
thickness,  the    proper   way   to 
try  4  for  that  thickness  is,  to 
multiply  it  by  3  times  20,  then 
by  itself,  and,  adding  the  prod- 
ucts to  1200,  multiply  the  sum  Fig.  2. 
by  4 ;  thus,  1200   sq.   in.  +  80 

sq.  in.  X3  +  16  sq.  in.  =  1456  sq.  in.,  and  1456X4  =  5824,  which 


\ 

0 

0 

0 

B 

B 

B 

p 

shows  that  4  in.  is  the  true  thickness,  and  20  in.  +  4  in.  —  the  re- 
quired edge,  24  in.,  Ans. 

NOTE.— Fig.  1  shows  also  the  complete  cube  with  the  section  lines 
marked. 


356 


RAY'S  HIGHER  ARITHMETIC. 


REMARKS. — 1.  Since  3  times  the  square  of  2  tens  is  equal  to  300 
times  the  square  of  2,  it  is  allowable  to  use  simply  the  square  of  the 
first  part  of  the  root  as  units,  multiplying  by  300  for  a  trial  divisor ; 
and  so,  too,  in  the  second  part  of  the  trial  work,  it  will  answer  to 
multiply  the  first  part  by  the  last  found  figure  and  by  30. 

2.  The  steps  are  trial 
steps,  and  as  remarked 
under  the  rule  for  square 
root,  our  artifices  have 
simply  narrowed  the  22X300  =  12 

2X4X30=    240 
42=       16 


OPERATION. 

14172488(242 


range  of  the  trial. 

3.  Had  the  power  been 
14172488,  there  would 
have  been  three  figures 
in  the  root ;  and  here,  as 
in  the  former  case,  the 
second  figure  is  4 ;  and 
we  may  treat  the  24  tens 
exactly  as  we  treated  the 
2  tens  in  the  former  illustration. 


1456 


6172 


5824 


24X^X30= 

22  = 


1440 
4 


174244 


348488 


348488 


SECOND  EXPLANATION. 

We  have  seen  (page  346)  how  the  cube  of  a  number,  of  two  figures, 
is  composed  ;  that,  for  example, 


Here  we  see  that  if  the  cube  of  the  tens  be  taken  away,  there  will 
remain 


that  is,  the  remainder  may  be  taken  as  a  product  of  two  factors,  of 
which  the  smaller  is  the  units,  and  the  larger  made  up  of  3  times  the 
square  of  the  tens,  with  3  times  the  tens  by  the  units,  with  also  the 
square  of  the  units.  Suppose,  then,  in  seeking  the  cube  root  of  74088, 
we  find  the  tens  to  be  4;  the  remainder  10088  must  be  the  product  of 
units  by  a  factor  composed  of  three  parts,  such  as  we  have  described. 
If  we  knew  the  larger  factor,  the  units  could  be  obtained  by  direct 
division  ;  but  we  do  know  that  larger  factor  to  be  greater  than  3  times 
the  square  of  40  ;  hence,  we  know  the  units  must  be  less  than  the 
quotient  of  10088  by  4800,  and  consequently  2  is  the  largest  figure  we 
need  try  for  units.  The  way  to  try  2,  is  to  compose  a  larger  factor 


EVOLUTION.  357 

after  the  manner  just  described  ;  OPERATION. 

hence,   multiply  2  by  3   times  74088(42 

40,  add  22  or  4,  add  the  sum  to  64 

the  4800,    and    multiply  by  2,  42X300  =  4800  10088 


which,  being  done,  shows  that        4  x  2X30—    240 
2  is  the  units  figure,  and  that  2  2  =          4 

42  is  the  root  sought.  5044  10088 


KEMARK. — Three  times  the  square  of  the  tens  is  the  convenient 
trial  divisor.  This  is  in  most  instances  a  greater  part  of  the  com- 
plete divisor ;  for  example,  the  least  number  of  tens  above  one  ten 
is  2,  and  the  greatest  figure  in  unit's  place  can  not  exceed  9;  the 
cube  of  29  is  24389,  the  first  complete  divisor  is  1821,  the  first  trial 
divisor  being  1200,  a  greater  part  of  it. 

378.  To  extract  the  cube  root  of  a  number  written 
in  the  decimal  notation  as  whole  number,  fraction, 
or  mixed  number. 

Rule. — 1.  Beginning  with  units,  separate  the  number  into 
periods  of  three  figures  each;  the  extreme  left  period  may  have 
but  one  or  two  figures,  but  the  extreme  right,  whether  of  units 
or  decimal  orders,  must  have  three  places,  by  the  annexing  of 
ciphers  if  necessary. 

2.  Find  the  greatest  cube  in  the  highest  period,  place  its  root 
on  the  right  as  a  quotient   in  division,  and  then   subtract  the 
cube  from  the  period,  bringing  down  to  the  right  of  the  remain- 
der the  next  period  to  complete  a  dividend. 

3.  Square  the  root  found  as  if  it  were  units,  multiply  it  by 
300,  and  place  the  product  on  the  left  as  a  trial  divisor ;  find 
how  often  it   is  contained  in  the  dividend,  and  place  the  quo- 
tient figure  to  the  right  of  the  root ;  multiply  the  quotient  by  30 
times  the  preceding  part   of  the  root  as  units,  square  also   the 
quotient,  and  add  the  tivo  results  to  the  trial  divisor;  then  mul- 
tiply the  sum   by  the  quotient,  and   subtract   the  product  from 
the    dividend,   annexing    to    the    remainder    another  period    as 
before. 

4.  Square  the  whole  root  as   before,   multiply  by  300,  pro- 


358  RAY'S  HIGHER  ARITHMETIC. 

ceeding  as  with  the  former  trial  divisor,  quotient,  and  addi- 
tions; continue  the  operation  until  the  remainder  is  nothing,  or 
until  the  lowest  required  decimal  order  of  the  root  has  been 
obtained. 

NOTES. — 1.  Should  any  product  exceed  the  dividend,  the  quotient 
figure  is  too  large. 

2.  If  any  remainder  is  larger  than  the  previous  divisor,  it  does 
not  follow  that  the  last  quotient  figure  is  too  small,  unless  the  remainder 
is  large  enough  to  contain  3  times  the  square  of  that  part  of  the  root  already 
found,  with  3  times  that  part  of  the  root,  and  1  more  ;  for  this  is  the  proper 
divisor  if  the  root  is  increased  by  1. 

3i  Should  decimal  periods  be  required  beyond  those  with  which 
the  operation  begins,  the  operator  may  annex  three  ciphers  to  each 
new  remainder. 

4.  When  the  operator  has  obtained  one  more  than  half  the  re- 
quired decimal  figures  of  the  root,  the  last  complete  divisor  and  the 
last  remainder  may  be  used  in  the  manner  of  contracted  division. 

379.  The  cube  root  of  any  product  is  equal  to  the  prod- 
uct of  the  cube  roots  of  the  factors.     Thus, 

1^250  X  4  X  648  X  9  —  ^125  X  8  X  216  X  27  = 
5X2X6X3. 

380.  The  cube  root  of  a  common  fraction  is  equal  to  the 
quotient  of  the  cube  root  of  the  numerator  by  the  cube  root 
of  the  denominator. 

REMARKS. — 1.  When  the  terms  are  not  both  perfect  cubes,  multi- 
ply both  by  the  square  of  the  denominator,  or  by  some  smaller  factor 
which  will  make  the  denominator  a  cube. 

2.  Reduce  mixed  numbers  to  improper  fractions,  or  the  fractional 
parts  of  such  numbers  to  decimals. 


EXAMPLES  FOR  PRACTICE. 


1.  f  512.  8. 

2.  1^19683.  27. 


3.  i3/7301384.  194. 

4.  if  94818816.  456. 


EVOLUTION.  359 


5.  if  1067462648.       1022. 

12.  if  25. 
13.  if  IT. 

2.924018 
2.22398 
.87358 
.64366 

6.  if  5.088448.             1.72 

7.  if  22188.041            28.1 

14.  iff- 

8.  if  32.65            3.196154 

15.  if  £. 

9.  if  .0079            .1991632 

16.  if  171 

.416328875     5.555 

10.  if  3.0092          1.443724 
11.  if^ST               .315985 

17.  if  7011.          19.1393267 

18.  if-^nHhf.              .2218845 
.6235319 

19.  if  |  of  T4T. 

EXTKACTION  OF   ANY  BOOT. 

381.  We  have  seen  in  the  chapter  on  Involution,  that 
if  a  power  be  raised  to  a  power,  the  new  exponent  is  the 
product  of  the  given  exponent  by  the  number  of  times  the 
power  is  taken  as  a  factor;  that,  for  example,  23  raised  to 
the  4th  power,  is  23X4—  212.  Consequently,  reversing 
that  process,  a  power  may  be  separated  into  equal  factors, 
if  the  given  exponent  be  a  composite  number;  thus,  212 
=  24X3,  and  consequently  i/2™  =  23,  orf  2TT^=24;  for 
212  equals  either  24  X  24  X  24,  or  23  X  23  X  23  X  23. 

It  is  important  to  note  the  distinction  between  separating  a  power 
into  factors  which  are  powers,  and  separating  that  power  into  equal 
factors,  having  the  same  roots  and  exponents.  The  latter  separa- 
tion would  be  extracting  a  root.  Thus, 

25=23X22  because  equal  to  23+2. 

But  the  square  root  of  25  is  not  23,  nor  is  the  cube  root  of  25  equal  to  22. 
If  23  be  taken  twice  as  a  factor  we  have  26,  and  1/26  =  23  ;  similarly, 


382.  A  root  of  any  required  degree  may  be  extracted,  by 
separating  the  number  denoting  the  degree,  into  its  factors,  and 
extracting  successively  the  roots  denoted  by  those  factors.  Thus, 
the  9th  root  is  the  cube  root  of  the  cube  root,  and  the  6th 
root  the  square  root  of  the  cube  root. 


360  RAY'S  HIGHER  ARITHMETIC. 


HORNEK'S  METHOD. 

383.  Horner's  Method,  named  from  its  inventor,  Mr. 
"W.  G.  Homer,  of  Bath,  England,  may  be  advantageously 
applied  in  extracting  any  root,  especially  if  the  degree  of 
the  root  be  not  a  composite  number. 

Rule  for  Extracting  any  Root. — 1.  Make  as  many 
columns  as  there  are  units  in  the  index  of  the  root  to  be 
extracted;  place  the  given  number  at  the  head  of  the  right- 
hand  column,  and  ciphers  at  the  head  of  the  others. 

2.  Commencing  at  the  right,  separate  the  given  number  into 
periods  of  as  many  figures  as  there  are  columns ;  extract  the 
required  root  to  within  unity,  of  the  left-hand  period,  for  the 
1st  figure  of  the  root. 

3.  Write  this  figure  in  the  1st  column,  multiply  it  then  by 
itself,  and  set  it  in  the  2d  column;  multiply  this  again  by 
the  same  figure,  and  set  it  in  the  3d   column,  and  so  on, 
placing  the  last  product  in  the  right-hand  column,  under  that 
part  of  the  gwen  number  from  which  the  figure  was  derived, 
and  subtracting  it  from  the  figures  above  it. 

4.  Add  the  same  figure  to  the  1st  column  again,  multiply 
the  result  by  the  figure  again,  adding  the  product  to  the  2d 
column,  and  so  on,  stopping  at  the  next  to  the  last  column. 

5.  Repeat  this  process,  leaving  off  one  column  at  the  right 
every  time,  until  all  the  columns  have  been  thus  dropped; 
then  annex  one  cipher  to  the  number  in  the  1st  column,  two 
to  the  number  in  the  2d  column,  and  so  on,  to  the  number  in 
the  last  column,  to  which  the  next  period  of  figures  from  the 
given  number  must  be  brought  down. 

6.  Divide  the  number  in  the  last  column  by  the  number  in 
the  previous  column  as  a  trial  divisor  (making  allowance  for 
completing  the  divisor^) ;  this  will  give  the  2d  figure  of  the  root, 
which  must  be  used  precisely  as  the  1st  figure  of  the  root  has 
been ;  and  so  on,  till  all  the  periods  have  been  brought  down. 


EVOLUTION.  361 


PROBLEM.— Extract  the  fourth  root  of  68719476736. 


0 
5 
5 
10 
5 

0 
25 

50 

0 
125 
375 

68719476736(512  Am. 
625 

75 
75 

*  500000 
15201 

*  621947 
515201 

15 
5 

*15000 
201 

515201 
15403 

*  1067466736 
1067466736 

*200 

15201 
202 

*530G04000 
3129368 

201 

15403 
203 

533733368 

202 
1 

*1560600 
4084 

203 
1 

1564684 

*2040 

2042 

NOTE. — It  is  convenient  to  denote  by  *  the  place  where  a  column  is 
dropped;  i.  e.,  reached  for  the  last  time  by  the  use  of  the  root 
figure  in  hand. 

384.     The  process  may  often  be  shortened  by  this 

Contracted  Method. — Obtain  one  less  than  half  of  the 
figures  required  in  the  root  as  the  rule  directs;  then,  instead 
of  annexing  ciphers  and  bringing  down  a  period  to  the  last 
numbers  in  the  columns,  leave  the  remainder  in  the  right-hand 
column  for  a  dividend;  cut  off  the  right-hand  figure  from  the 
last  number  of  the  previous  column,  tivo  right-hand  figures 
from  the  last  number  in  the  column  before  that,  and  so  on, 
always  cutting  off  one  more  figure  for  every  column  to  the  left. 

With  the  number  in  the  right-hand  column  and  the  one 
in  the  previous  column,  determine  the  next  figure  of  the  root, 
and  use  it  as  directed  in  the  ride,  recollecting  that  the  figures 
cut  off  are  not  used  except  in  carrying  the  tens  they  produce. 

H.  A.  31. 


362  KAY'S  HIGHER  ARITHMETIC. 

This  process  is  continued  until  the  required  number  of  figures 
is  obtained,  observing  that  when  all  the  figures  in  the  last 
number  of  any  column  are  cut  off,  that  column  ivill  be  no 
longer  used. 

REMARK.  —  Add  to  the  1st  column  mentally;  multiply  and  add 
to  the  next  column  in  one  operation:  multiply  and  subtract  from 
the  right-hand  column  in  like  manner. 

PROBLEM.  —  Extract  the  cube  root  of  44.6  to  six  decimals. 

0  0  44.600(3.546323 

3  9  17600 

6  2700  1725000 

90  3175  238136 

95  367500  12182 

100  371716  865 

1050  37594$  111 

1054  37659 

1058 


REMARK.  —  The  trial  divisors  may  be  known  by  ending  in  two 
ciphers;  the  complete  divisors  stand  just  beneath  them.  After  get- 
ting 3  figures  of  the  root,  contract  the  operation  by  last  rule. 

EXAMPLES  FOR  PRACTICE. 

1.  Extract  the  square  root  of  15625.  125. 

2.  Extract  the  cube  root  of  68719476736.  4096. 

3.  Extract  the  fifth  root  of  14348907. 

4.  The  cube  root  of  151.  5.325074 

5.  J/9TAL  3.1416 

6.  ^L08  1.01943 

7.  T^S-  .83938 

8.  j/35^~  2.03848 

9.  ^782757789696.  9.79795897 
10.  ^1367631.  4.8058955 


EVOLUTION. 


363 


APPLICATIONS  OF  SQUARE   ROOT  AND  CUBE  ROOT. 
DEFINITIONS 

385.     1.  A  Triangle    is  a  figure  which    has  three  sides 
and  three  angles;  as,  ABC,  MNP. 


2.  The  Base  of  a   triangle   is  that  side  upon  which  it  is 
supposed  to  stand ;  as,  AB,  MN. 

3.  The  Altitude  of  a   triangle  is  the   perpendicular  dis- 
tance   from   the   base  to  the  vertex  of  the  angle  opposite ; 
as,  HP. 

REMARK. — The  three  angles  of  a  triangle  are  together  equal  to 
180°,  or  two  right  angles.  The  proof  of  this  belongs  to  Geometry, 
but  a  fair  illustration  may  be  made  in  the  manner  indicated  above. 
Mark  the  angles  of  a  card  or  paper  triangle,  1,  2,  3;  and  by  two  cuts 
divide  it  into  three  parts.  Place  the  marked  angles  with  their  ver- 
tices as  at  O,  and  it  will  be  seen  that  the  pieces  fit  a  straight  edge 
through  O,  while  the  angles  cover  just  twice  90°,  or  EOD  +  FOD. 
Any  angle  less  than  90°,  as  HOF,  is  an  acute  angle;  any  angle  greater 
than  90°,  as  HOE,  is  an  obtuse  angle. 

4.  An  Equilateral  Triangle  is  a  triangle  having    three 
equal  sides ;  as,  MNP. 

REMARK. — The  angles  of  an  equilateral  triangle  are  60°  each ; 
hence,  six  equilateral  triangles  can  be  formed  about  the  same  point 
as  a  vertex,  each  angle  at  the  vertex  being  measured  by  the  sixth  of  a 
circumference.  (Art.  204.) 


364 


It  AY'S  HIGHER  ARITHMETIC. 


EIGHT-ANGLED  TRIANGLES. 

388.  A  Right-angled  Triangle  is  a  triangle  having 
one  right  angle ;  as  IGK,  where  G  =  90°. 

The  side  opposite  the  right  angle  is  called 
the  hypothenuse ;  the  other  two  sides  are  called 
the  base  and  perpendieular. 

It  is  demonstrated  in  Geometry  that  the 
square  described  upon  the  hypothenuse  of  a 
right-angled  triangle  is  equal  to  the  sum  of 
the  squares  described  on  the  other  two  sides. 

ILLUSTRATION. — A  practical  proof  of  this  may  be  made  in  the 
following  manner,  especially  valuable  when  the  triangle  has  no 
equal  sides.  It  will  be  a  useful  and  entertaining  exercise  for  the 
pupil. 

Let  the  triangle  be  described  upon  a  card,  and  let  it  stand  upon 
the  hypothenuse,  as  AEB  does.  Make  three  straight  cuts ;  —  one, 
perpendicular  from  A,  through  the  smaller  square ;  one,  perpendic- 
ular from  B,  through  the  larger  square,  and  one  at  right  angles 
from  the  end  of  the  second  cut.  The  two  squares  are  thus  divided 


into  five  parts,  which  may  be  marked,  and  arranged,  as  here  shown, 
in  a  square  equal  to  one  described  on  the  hypothenuse. 

REMARK. — The  perpendicular  in  an  equilateral  triangle  divides 
the  base  into  two  equal  parts,  and  also  divides  the  opposite  angle  into 
two  which  are  80°  each.  From  this  it  follows  that  if,  in  a  right- 


EVOLUTION.  365 

angled  triangle,  one  angle  is  30°,  the  side  opposite  that  angle  is  half 
the  hypothenuse  ;  and,  conversely,  if  one  side  be  half  the  hypothenuse, 
the  angle  opposite  will  be  30°. 

387.  To  find  the  hypothenuse  when  the  other  two 
sides  are  given. 

Rule.  —  Add  together  the  squares  of  the  base  and  perpendicular, 
and  extract  the  square  root  of  the  sum. 

388.  To   find   one  side  when  the  hypothenuse  and 
the  other  side  are  given. 

Rule.  —  Subtract  the  square  of  the  given  side  from  the  square 
of  the  hypotiienuse,  and  extract  the  square  root  of  Hie  differ- 
ence; or, 

Multiply  the  square  root  of  ilie  sum  of  the  hypotJienuse  and 
side  by  the  square  root  of  their  difference. 

Representing  the  three  sides  by  the  initial  letters  h,  p,  b,  we  have 
the  following 

FORMULAS.  —  1.  h  =  Vp*  +  b^. 


2.  p  =  Vh2—b*  ;  or,  p  =  Vh  +  b  X  v'T^—b. 

3.  b  =  Vh2  —p2  ;  or,  b  =  Vh-^-p  X  Vh—  p. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  length  of  a  ladder  reaching  12  ft.  into  the 
street,  from  a  window  30  ft,  high.  32.31  +  ft. 

2.  What  is  the  diagonal,  or  line  joining  the  opposite  cor- 
ners, of  a  square  whose  side  is  10  ft.  ?  14.142  -f-  ft. 

3.  What  is  saved  by  following  the  diagonal  instead  of  the 
sides,  69  rd.  and  92  rd.,  of  a  rectangle?  46  rd. 

4.  A  boat  in  crossing  a  river  500  yd.  wide,  drifted  with 
the  current  360  yd. ;  how  far  did  it  go?  616  +  yd. 


366  RAY'S  HIGHER  ARITHMETIC. 

REMARK. — Integers  expressing  the  sides  of  right-angled  triangles 
may  be  found  to  any  extent  in  the  following  manner :  Take  any  two 
unequal  numbers ;  the  sum  of  their  squares  may  represent  a  hypoth- 
enuse ;  the  difference  of  their  squares  will  then  stand  for  one  side,  and 
double  their  product  for  the  remaining  side.  Thus,  from  3  and  2,  form 
13,  12,  5 ;  from  4  and  1,  form  17,  15,  8 ;  from  5  and  2,  form  29,  21,  20. 


PARALLEL  LINES  AND  SIMILAR  FIGURES. 
DEFINITIONS. 

389.  1.  Parallel  Lines  are  lines  which  have  the  same 
direction.  The  shortest  distance  between  two  straight  par- 
allels is,  at  all  points,  equal  to  the  same  perpendicular  line. 

2.  Similar  Figures  are  figures  having  the  same  number 
of  sides,  and  their  like  dimensions  proportional. 

REMARKS. — 1.  Similar  figures  have  their  corresponding  angles 
equal. 

2.  If  a  line  be  drawn  through  any  triangle  parallel  to  one  of  the 
sides,  the  other  two  sides  are  divided  proportionally,  and  the  triangle 
marked  off,  is  similar  to  the  whole  triangle.     An  illustration  of  this 
has  already  been  furnished  in  solving  Ex.  8,  Art.  231. 

3.  All  equilateral  triangles  are  similar;   the  same  is  true  of   all 
squares,  all  circles,  all  spheres. 

3.  The  areas  of  similar  figures  are  to  each  other  as  the 
squares  of  their  like  dimensions. 

4.  The  solidities  of  similar  solids  are  to  each  other  as  the 
cubes  of  their  like  dimensions. 


GENERAL    EXERCISES   IN   EVOLUTION   AND   ITS 
APPLICATIONS. 

1.  One  square  is  12 \  times  another:  how  many  times  does 
the  side  of  the  1st  contain  the  side  of  the  2d  ?  3  J. 

2.  The  diagonals  of  two  similar  rectangles  are  as  5  to  12 : 
how  many  times  does  the  larger  contain  the  smaller? 


EVOLUTION.  367 

3.  The  lengths  of  two  similar  solids  are  4  in.  and  50  in. ; 
the  1st  contains  16  cu.  in.:  what  does  the  2d  contain? 

31250  cu.  in. 

4.  The  solidities  of  two  balls  are  189  cu.  in.  and  875  cu. 
in. ;  the  diameter  of  the  2d  is  17^  in.  ;  find  the  diameter  of 
the  1st.  10^  in. 

5.  In  extracting  the  square  root  of  a  perfect  power,  the 
last    complete   dividend   was    found    4725:    what   was    the 
power?  225625. 

6.  What  number  multiplied  by  f  of  itself  makes  504? 

42. 

7.  Separate  91252^  into   three  factors  which  are  as  the 
numbers  1,  2A,  and  3*1  23,  57.5,  and  69. 

8.  What  integer  multiplied  by  the  next  greater,  makes 
1332  ?  36. 

9.  The  length  and  breadth  of  a  ceiling  are  as  6  and  5 ; 
if  each  dimension  were  one  foot  longer,  the  area  would  be 
304  sq.  ft. :  what  are  the  dimensions?  18  ft.,  15  ft. 

10.  In  extracting    the  cube   root  of  "a    perfect    integral 
power,  the  operator  found  the  last  complete  dividend  241984: 
what  was  the  power?  2985984. 

11.  If  we  cut  from  a  cubical  block  enough  to  make  each 
dimension  one  inch  shorter,  it -will  lose  1657  cubic  inches: 
what  is  the  solidity?  13824  cu.  in. 

12.  A  hall  standing  east  and  west,  is  46  ft.  by  22  ft.,  and 
12^  ft.  high :  what  is  the  length  of  the  shortest  path  a  fly 
can  travel,  by  walls  and  floor,  from  a  southeast  lower  corner 
to  a-  northwest  upper  corner  ?  57-|  ft. 

13.  How  many  stakes  can  be  driven  down  upon  a  space 
15  ft.  square,  allowing  no  two  to  be  nearer  each  other  than 
li  ft.,  and  how  many  allowing  no  two  to  be  nearer  than  1J 
ft?  128,  and  180  stakes. 

14.  What  integer  is  that  whose  square  root  is  5  times  its 
cube  root?  15625. 

15.  If  the   true  annual   rate  of  interest  be    10%,  what 
would  be  the  true  rate  for  each  73  days,  if  the  interest  be 


368 


RAY'S  HIGHER  ARITHMETIC. 


compounded  through  the  year?  Prove  the  result  by  con- 
tracted multiplication.  (Art.  334,  Kern.  2.)  1.924%. 
16.  If  a  field  be  in  the  form  of  an  equilateral  triangle 
whose  altitude  is  4  rods,  what  would  be  the  cost  of  fencing 
it  in,  at  75  ct.  a  rod?  $10.39 


1.  Involution..., 


2.  Evolution  . 


Topical  Outline. 


POWERS  AND  ROOTS. 


1.  Definitions. 


2.  Terms.. 


3.  Squaring  and 

Cubing.... 


1.  Definitions. 


2.  Terms, 


1.  Powers. 

2.  Root. 

3.  Degree. 

.  4.  Exponent. 

1.  Algebraic  Statements. 

2.  Numerical  Illustrations. 

3.  Geometrical  Illustration. 
.  4.  Principles. 


Perfect. 
Imperfect. 


1.  First  explanation 

(Geometrical). 

3.  Square  Root -j   2.  Second  explanation 

(Algebraic). 
[  3.  Rule. 

1.  First  explanation 

(Geometrical). 

4.  Cube  Root -j   2.  Second  explanation 

(Algebraic). 
3.  Rule. 

5.  Roots  in  General — Homer's  Method. 

6.  Applications. 


XXI.   SEEIES. 

DEFINITIONS. 

390.  1.  A  Series  is   any  number  of  quantities   having 
a  fixed  order,  and  related  to  each  other  in  value  according 
to  a  fixed  law.     These  quantities  are  called  Terms;  the  first 
and  last  are  called  Extremes,  and  the  others  Means. 

The  Law  of  a  series  is  a  statement  by  which,  from  some  necessary 
number  of  the  terms  the  others  may  be  computed. 

2.  There  are  many  different  kinds  of  series.  Those  usu- 
ally treated  in  Arithmetic  are  distinguished  as  Arithmetical 
and  Geometrical;  these  series  are  commonly  called  Progres- 
sions. 

AKITHMETICAL    PROGKESSION. 

391.  1.  An   Arithmetical    Progression  is   a  series  in 
which    any   term    differs    from    the    preceding   or   following 
by  a  fixed  number.     That  fixed  number  is  called  the  com- 
mon difference;  and  the  series  is  Ascending  or  Descending, 
accordingly  as  the  first  term  is  the  least  or  the  greatest. 

Thus,  1,  3,  5,  7,  9,  is  an  ascending  series,  whose  common  differ- 
ence is  2 ;  but  if  it  were  written  in  a  reverse  order  (or,  if  we  treated 
9  as  the  first  term),  the  series  would  be  descending. 

2.  Every  Arithmetical  Progression  may  be  considered 
under  the  relations  of  five  quantities,  such  that  any  three 
of  them  being  given,  the  others  may  be  found.  These  five  are 
conveniently  represented  as  follows: 

First  term, a. 

Last  term, I. 

Number  of  terms, .     .     .     .     n. 

Common  difference,  d. 

Sum  of  all  the  terms,     .     .      s. 

(369) 


370  RAY'S  HIGHER  ARITHMETIC. 

3.  These  give  rise  to  twenty  different  cases,  but  all  the 
calculations  may  be  made  from  the  principles  stated  in  the 
two  following  cases. 

NOTE. — Some  of  the  problems  arising  under  this  subject  are,  prop- 
erly, Algebraic  exercises.  Nothing  will  be  presented  here,  however, 
which  is  beyond  analysis  by  means  of  principles  and  processes  exhib- 
ited in  this  book.  The  formulas  given  are  easily  understood,  and  the 
student  will  find  it  a  very  simple  operation  to  write  the  numbers 
in  place  of  their  corresponding  letters,  and  work  according  to  the 
signs.  The  formulas  are  presented  as  a  convenience. 

CASE   I. 

392.  One  extreme,  the  common  difference,  and  the 
number  of  terms  being  given,  to  find  the  other  ex- 
treme. 

PROBLEM. — Find  the  20th  term  of  the  arithmetical  series 
1,  4,  7,  10,  etc. 

SOLUTION. — Here  the  series  may  be  considered  as  made  of  20 
terms,  and  we  seek  the  last.  The  com.  din",  is  3,  and  the  terms 
are  composed  thus:  1,  1  +  3,  1+6,  1  +  9,  etc.;  and  it  is  obvious 
that,  as  the  addition  of  the  com.  difF.  commences  in  forming  the  second 
term,  it  is  taken  twice  in  the  third  term,  three  times  in  the  fourth, 
and  so  on;  similarly  therefore  it  must  be  taken  19  times  in  form- 
ing the  20th,  and  the  simple  operation  is,  1  +  (20  —  1)X3  —  58,  Ans. 

FORMULA. — I  =  a  +  (n  —  I)c7;  or  I  —  a  —  (n  —  l)d. 

Rule. — Multiply  the  number  of  terms  less  one  by  the  com- 
mon difference,  add  the  product  to  the  given  extreme  when  the 
larger  is  sought ,  subtract  it  from  the  given  extreme  when  the 
smaller  is  sought. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  12th  term  of  the  series  3,  7,  11,  etc.         47. 

2.  Find  the  18th  term  of  the  series  100,  96,  etc.         32. 

3.  Find  the  64th  term  of  the  series  3|,  5f ,  etc.         145J. 


SERIES.  371 

4.  Find  the  10th  term  of  the  series  .025,  .037,  etc.     .133 

5.  Find   the   1st   term  of  the  series   68,   71,  74,  having 
19  terms.  20. 

6.  Find  the  1st  term  of  the  series  117,  123^,  130,  having 
6  terms.  97J. 

7.  Find  the  first  term  of  the  series  18f  ,  12|,  6£,  having 
365  terms.  2281J. 

CASE     II. 

393.     The  extremes  and  the  number  of  terms  being 
given,  to  find  the  sum  of  the  series. 

PROBLEM.  —  What  is  the  sum  of  9  terms  of  the  series  1, 

4,  7,  10,  etc.? 

OPERATION. 

EXPLANATION.  —  Writing  the  series         14-  (9  _  1  )  X  3  —  25 
in   full,  in   the   common   order,  and 

also  in  a  reverse  order,  we  have  "j"       \/  9  __  jjy    ^^ 

2 


Sum  =  1+4  +  7  +  10  +  13  +  16  +  19 

Sum  =  25  +  22  +  19  +  16  +  13  +  10  +  7  +  4  +  1. 

Twice  the  sum  =  26  +  26  +  26  +  26  +  26  +  26  +  26  +  26  +  26  =  9 
times  the  sum  of  the  extremes;  /.  the  sum  —  J  of  9X26  —  117,  Am. 
If  we  add  a  term  whose  place  is  a  certain  distance  beyond  the 
first,  to  another  whose  place  is  equally  distant  from  the  last,  the 
sum  will  be  the  same  as  that  of  the  extremes,  and  hence,  as  the 
above  illustrates,  the  double  of  any  such  series  is  equal  to  the 
product  of  the  number  of  terms  by  the  sum  of  the  extremes. 

FORMULA.  —  s  —  (a  +  /)  n. 

~2~ 

"Rule.—  Multiply   the   sum  of  the  extremes    by  the    number 
of  terms,  and  divide  by  2. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  sum  of  the  arithmetical  series  whose  extremes 
are  850  and  0,  and  number  of  terms,  57.  24225. 


372  RAY'S  HIGHER  ARITHMETIC. 

2.  Extremes,    100  and  .0001:   number  of  terms,   12345. 

617250.61725 

3.  What  is  the  sum  of  the  arithmetical  series   1,  2,  3, 
etc.,  having  10000  terms?  50005000. 

4.  Of  1,  3,  5,  etc.,  having  1000  terms?  1000000. 

5.  Of  999,  888,  777,  etc.,  having  9  terms?  4995. 

6.  Of  4.12,  17.25,  30.38,  etc.,  having  250  terms? 

409701.25 

7.  Whose  5th  term  is  21;    20th   term,   60;    number  of 
terms,  46?  3178f. 

EXAMPLES  FOR  PRACTICE. 

REMARK. — It  is  not  deemed  necessary  to  formulate  a  special  rule 
for  each  class  of  examples  here  introduced.  The  following  are  pre- 
sented as  exercises  in  analysis,  each  depending  on  one  or  more  of 
the  principles  above  stated. 

1.  Find  the  common  difference  of  a  series  whose  extremes 
are  8  and  28,  and  number  of  terms,  6.  4. 

2.  Extremes   are    4^    and    20f,    and    number   of  terms, 
14.  1J. 

3.  Insert  one  arithmetical  mean  between  8  and  54.        31. 

4.  Insert  five  arithmetical  means  between  6  and  30. 

10,  14,  18,  22,  26. 

5.  Insert  two  arithmetical  means  between  4  and  40. 

16,  28. 

6.  Insert  four  arithmetical  means  between  2  and  3. 

21,  2|,  2|,  2f 

7.  What  is  the  number  of  terms  in  a  series  whose  ex- 
tremes are  9  and  42,  and  common  difference,  3?  12. 

8.  Whose  extremes  are  3  and   10^,  and  common  differ- 
ence, §?  21. 

9.  In  the  series  10,  15  ...  500?  99. 

10.  What  principal,  on  annual  interest  at  10^,  will,  in 
50  yr.,  amount  to  $4927.50?  $270. 


SERIES.  373 


GEOMETRICAL    PROGRESSION. 

394.  1.  A  Geometrical  Progression  is  a  series  in  which 
any  term  after  the  first  is  the  product  of  the  preceding  term 
by  a  fixed  number.     That  fixed  number  is  called  the  ratio; 
and  the  series  is  ascending  or  descending  accordingly  as  the 
first  term  is  the  least  or  the  greatest. 

Thus,  1,  3,  9,  27,  is  an  ascending  progression,  and  the  com- 
mon multiplier  is  the  ratio  of  3  to  1  (Art.  228,  2),  or  of  any 
term  to  the  preceding;  considering  27  as  the  first  term,  the  same 
series  may  be  called  descending. 

2.  Any  Geometrical  Progression  may  be  considered  under 
the  relations   of  five  quantities,  of  which    three    must    be 
known  in   order  to   find   the  others.      These  five  are  thus 
represented : 

First  term, a. 

Last  term,    ...'.../. 

Ratio, r. 

Number  of  terms,  .     .     .     .     n. 
Sum  of  all  the  terms,     .     .     s. 

3.  These  quantities   give   rise   to  20   different  classes  of 
problems,  but  all  of  the  necessary  calculations  depend  upon 
principles  set  forth  in  the  following  cases. 

NOTE. — Some  of  the  problems  arising  from  these  quantities  require 
such  an  application  of  the  formulas  as  can  not  be  understood  without 
a  knowledge  of  Algebra. 

CASE   I. 

395.  From    one    extreme,   the    common    ratio    and 
the   number   of  terms,   to   find  the   other   extreme. 

PROBLEM.— Find  the  8th  term  of  the  series  1,  2,  4, 
8,  etc. 


374  RAY'S  HIGHER  ARITHMETIC. 

EXPLANATION. — Here,   1   being  the  first  STATEMENT. 

term,  and  2  the  ratio,  we  see  that  the  series  27  X  1  =  1  2  8,  Ans. 

may    be    formed    thus:    1,    1X2,    1  X  22, 

1X23,  etc.,  the  ratio  being  raised  to  its  second  power  in  forming 
the  3d  term,  to  its  third  power  in  forming  the  4th ;  and  so,  similarly, 
the  8th  term  =  1X27  =  128,  Ans. 

FORMULA. — I  =  arn-l. 

Rale. — Consider  the  given  extreme  as  the  first  term,  and 
multiply  it  by  that  power  of  the  ratio  whose  degree  is  denoted 
by  the  number  of  terms  less  one. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  last  term  in  the  series  64,  32,  etc.,  of  12 
terms.  -g^. 

2.  In  2,  5,  12  J,  etc.,  having  6  terms.  195TV 

3.  In  100,  20,  4,  etc.,  having  9  terms.  15|25. 

4.  1st  term,  4;  common  ratio,  3;  find  the  10th  term. 

78732. 

5.  3d  term,  16;  common  ratio,  6:  find  the  9th  term. 

746496. 

6.  33d  term,  1024;  common  ratio,  f  :  find  the  40th  term. 

136H. 

7.  Find  the  1st  term  of  the  series  90,  180,  of  6  terms.    5f. 

8.  Of  ^_%>  IHfc  havin£  n  terms-  TT>  s- 


CASE    II. 

396.     Prom  one  extreme,  the  ratio,  and  the  number 
of  terms,  to  find  the  sum  of  the  terms. 

PROBLEM.  —  The  first  term  is  3,  the  ratio  4,  the  number 
of  terms  5;  required  the  sum  of  the  series. 

OPERATION. 

EXPLANATION.—  Writing  the  4  X3X44  —  3 

.  -       -       —  =  1  u  L  6, 

whole  series,  we  have  :  4  —  1 


SERIES.  375 

S  =  3  +  12  +  48  +  192  +  768. 
Also,  4  S  =         12  +  48  +  192  +  768  +  3072. 

It  is  evident  that  the  lower  line  exceeds  the  upper  by  the  difference 
between  3072  and  3;  this  difference  may  be  written  4X768  —  3,N)r 
4  X  3  X  4  4  —  3,  and  as  this  is  3  times  the  series,  we  have,  once  the 
series  = 

/1X/OK//14.  O 

-=1023,  Ans. 


4—1 

Now,  observing  the  form,  note  that  we  have  multiplied  the  last  term 
by  the  ratio,  then  subtracted  the  first  term,  and  then  divided  by  the 
ratio  less  one.  If  the  series  had  stood  with  768  for  first  term,  and  the 
multiplier  £,  we  should  have  had 

768  +  192  +  48  +  12  +  3         =  S, 

1  Qf)        I         A  O        [        1  f)        |        O        |        3    1       Q  . 

]L\J—i  ~r~  TtO  ~T~  -L^  ~r~  O  — r~  x 2T    ^  > 

and  thus  768  —  f  =  f  of  the  series  ^  hence,  we  can  write, 
768  — j:X3^  x  0  2  3>  as  before> 

Here  we  have  taken  the  product  of  the  last  term  by  the  ratio  from  the 
first  term,  and  have  divided  by  the  excess  of  unity  above  the  ratio. 
In  either  case,  therefore,  we  have  illustrated 

Rule  I. — Find  the  last  tervi  and  multiply  it  by  the  ratio; 
then  find  the  difference  between  this  product  and  the  first  term, 
and  divide  by  the  difference  between  the  ratio  and  unity. 

a      rl  —  a  Q       a  —  rl 

FOKMULA. —  8= —  ;  or,  S  = . 

r  —  1  1  —  r 

The  first  answer,  above  given,  may  take  another  form,  thus : 

-   is  the  same  as    -,   where  appear   the 

4  —  1  4  —  1 

ratio  4,  the  first  term  3,  and  the  number  of  terms  5.  This  form,  often 
used  when  the  series  is  ascending,  has  the  following  general  statement : 

S  =  •£=!) 

and  corresponds  to  the  following  rule ; 


376  RAY'S  HIGHER  ARITHMETIC. 

Rule  II. — Raise  the  ratio  to  a  power  denoted  by  the  num- 
ber of  terms,  subtract  1,  divide  the  remainder  by  the  ratio  less  1, 
and  multiply  the  quotient  by  the  first  term. 

NOTES. — 1.  The  amount  of  a  debt  at  compound  interest  for  a  num- 
ber of  complete  intervals,  is  the  last  term  of  a  geometrical  progression, 
whose  ratio  is  1  +  the  rate  per  cent.  The  table  (Art.  335)  shows 
the  powers  of  the  ratio.  For  example,  the  period  being  4  yr.,  the 
number  of  terms  is  five;  the  first  term  is  the  principal,  and  the  power  of 
the  ratio  required  by  Case  I,  is  the  fourth. 

2.  The  amount  of  an  annuity  at  compound  interest  is  conveniently 
found   by  the  Formula  corresponding  to  Kule  II.     The   table  (Art. 
335)  is  available,  and  the  work  very  simple.     Thus,  if  the  annuity 
be  $200,  the  time  40  yr.,  and  the  rate  6^,,  we  have  a  =  200,  n  =  40, 
r  =  1.06     Then,  writing  these  values  in  the  Formula,  we  have : 

$20Q(1.0640— 1)      $200X9.2857179 

Amount  =  * -=—  ——$30952.39,  Ans. 

1.06  —  1  .06 

3.  If  the  series  be  a  descending  one  having  an  infinite  number  of 
terms,  the  last  term  is  0,  and  the  product  required  by  Rule  I  is  0. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  sum  of  6,  12,  24,  etc.,  to  10  terms.         6138. 

2.  Of  16384,  8192,  etc.,  to  20  terms.  32767ff 

3.  Of  f ,  |,  2-8T,  etc.,  to  7  terms.  liftf 
Find  the  sum  of  the  following  infinite  geometrical  series : 

4.  Of  1,  i    i,  etc.  2. 
5-  Of  f ,  A,  ^  etc.                                                        1|. 

6.  Of  £,  f ,  &,  etc.  2. 

7.  Of  £,  1,  f  etc.  81 

8.  Of  .36  =  .3636,  etc.  .=  T\6o  +  Trftfinr,  etc-  T4y 

9.  Of  .349206,  of  480,  of  6.  ff  and  £ff  and  f. 

10.  Find  the  amount  of  an  annuity  of  $50,  the  time  being 
53  yr.,  the  rate  per  cent  10.  $77623.61 

11.  Applying   the   formula  used   in  the  last  example,  to 
any  case  of  the  same  kind,  prove  the  truth  of  the  rule,  in 
Case  IV,  of  Annuities. 


SERIES.  377 

12.  Calculate  a  table  of  amounts  of  an  annuity  of  $1,  for 
any  number  of  years  from  1  to  6,  at  8^. 

REMARK. — It  is  not  considered  necessary  to  give  special  rules  for 
finding  the  ratio,  and  the  number  of  terms  when  these  are  unknown  ; 
so  far  as  these  are  admissible  here,  they  involve  no  principles  beyond 
what  are  presented  in  the  matter  already  given. 


EXAMPLES  FOR  PRACTICE. 

1.  Find   the  common  ratio:    first  term,  8;  fourth  term, 
512.  4. 

2.  First  term,  4}|;  eleventh  term,  49375000000.          10. 

3.  Sixteenth  term,  729  ;  twenty-second  term,  1000000.    3£. 

4.  Insert  1  geometric  mean  between  63  and  112.  84. 

5.  Four  geometric  means  between  6  and  192. 

12,  24,  48,  96. 

6.  Three  geometric  means  between  -^-g^,  and  -J-. 

TeVs'    5T6">    T2- 

7.  Two  geometric  means  between  14.08  and  3041.28 

84.48  and  506.88 


Topical  Outline. 
SERIES. 

f  1.  Definitions. —Terms,  Law,  Extremes.  Means. 

Series....  f  L  Terms- 

1.  Arithmetical...  J   2.  Cases. 

2.  Classes...  j  (Formulas.) 

f  1.  Terms. 
.  2.  Geometrical....  J   ^  Case& 

I        (Formulas.) 
H.  A.  32. 


XXII.  MENSUEATIOK. 

DEFINITIONS. 

397.  1.  Geometry  is  that  branch  of  mathematics  which 
treats   of  quantity   having    extension   and   form.     When   a 
quantity  is  so  considered,  it  is  called  Magnitude. 

2.  There  are   four  kinds  of  Magnitude  known  to  Geom- 
etry :  Lines,  Angles,  Surfaces,  and  Solids.     A  point  has  posi- 
tion, but  not  magnitude. 

3.  Mensuration  is  the  application  of  Arithmetic  to  Geom- 
etry ;  it  may  be  defined  also  as  the  art  of  computing  lengths, 
areas,  and  volumes. 

LINES. 

398.  1.  A  line  is  that  which  has  length  only. 

2.  A  straight  line  is  the  shortest  distance  between  two 
points. 

3.  A  broken  line  is  a  line  made  of  connected  straight 
lines  of  different  directions. 

4.  A  curve,  or  curved  line,  is  a  line   having  no   part 
straight. 

The  word  "line,"  used  without  the  qualifying  word  "curve,"  is  un- 
derstood to  mean  a  straight  line. 

5.  A  horizontal  line  is  a  line  parallel  with  the  horizon, 
or  with  the  water  level.     (See  Art.  389;  1.) 

6.  A  vertical  line  is  a  line  perpendicular  to  a  horizontal 
plane. 

(378) 


MENSURATION.  379 


ANGLES. 

399.     An   angle  is   the   opening,  or   inclination,   of  two 
lines  which   meet   at   a   point.     (Art.   204.) 

REMARK. — Angles   differing  from   right  angles   are   called   oblique 
angles.     (See  Art.  385,  3,  Rem.,  and  Art.  386.) 


SURFACES. 
POLYGONS. 

400.  1.  A  surface  is  that  which  has  length  and  breadth 
without  thickness. 

A  solid  has  length,  breadth,  and  thickness. 

A  line  is  meant  when  we  speak  of  the  side  of  a  limited  surface,  or 
the  edye  of  a  solid ;  a  surface  is  meant  when  we  speak  of  the  side  or 
the  base  of  a  solid. 

2.  A  Plane  is  a  surface  such  that  any  two  points  in   it 
can  be  joined    by  a  straight   line  which   lies  wholly  in  the 
surface.     The  application  of  a  straight-edge  is  the  test  of  a 
plane. 

3.  A  plane  figure  is  any  portion  of  a  plane  bounded  by 
lines. 

4.  A  polygon  is  a  portion  of  a  plane  inclosed  by  straight 
lines ;  the  perimeter  of  a  polygon  is  the  whole  boundary. 

5.  Area  is  surface  defined  in  amount.     For  the  numerical 
expression  of  area,  a  square  is  the  measuring  unit.     (Art. 
197.) 

6.  A  polygon  is  regular  when  it  has  all  its  sides  equal, 
and  all  its  angles  equal. 

7.  A  polygon  having  three  "sides  is  a  trian-          r^^\N 
gle ;    having   four  sides,   a  quadrilateral ;    five  V 
sides,  a  pentagon;  six  sides,  a  hexagon,  etc.          P  Quadrilateral. 


380  RA  YJS  HIGHER  ARITHMETIC. 

The  six  diagrams  following  represent  regular  polygons. 


Pentagon .          Hexagon . 


Heptagon. 


Octagon ., 


Nouagon. 


Decagon . 


8.  The  diagonal  of  a  polygon  is  the  straight  line  joining 
two  angles  not  adjacent ;  as,  PN,  on  the  preceding  page. 

9.  The  base  is  the  side  on  which  a  figure  is  supposed  to 
stand. 

10.  The  altitude  of  a  polygon  is  the  perpendicular  distance 
from  the  highest  point,  or  one  of  the  highest  points,  to  the 
line  of  the  base. 

11.  The  center  of  a  regular  polygon  is  the  point  within, 
equally  distant   from   the   middle   points  of  the   sides ;  the 
apothem  of  such  a  polygon  is  the  perpendicular  line  drawn 
from  the  center  to  the  middle  of  a  side ;  as,  C  a  center,  and 
CD  an  apothem. 

TRIANGLES. 

401.     Triangles  are  classified  with  respect' to  their  angles, 
and  also  with  respect  to  their  sides. 


Acute  Triansrles. 


Obtuse  Triangles. 


Scalene. 


Isosceles.         Equilateral.        Isosceles. 


1.  A  triangle  is  right-angled  when  it  has  one  right  angle  ; 
it  is  acute-angled  when  each  angle  is  acute;  it  is  obtuse- 
angled  when  one  angle  is  obtuse.  These  three  classes  may 
be  named  right  triangles,  acute  triangles,  obtuse  triangles ; 
the  last  two  classes  are  sometimes  called  oblique  triangles. 


MENSURATION.  381 

2.  A  triangle  is  scalene  when  it  has  no  equal  sides; 
isosceles,  when  it  has  two  equal  sides;  and  equilateral, 
when  its  three  sides  are  equal. 

A  right  triangle  can  be  scalene,  as  when  the  sides  are  3,  4,  5 ;  or, 
isosceles,  as  when  it  is  one  of  the  halves  into  which  a  diagonal  di- 
vides a  square.  An  obtuse  triangle  can  be  scalene  or  isosceles;  an 
acute  triangle  can  be  scalene,  isosceles,  or  equilateral. 


QUADRILATERALS. 

402.     Quadrilaterals  are  of  three  classes: 

1.  A  Trapezium  is  a  quadrilateral  having  no  two  sides 
parallel. 

2.  A  Trapezoid  is  a  quadrilateral  having  two  and  only 
two  sides  parallel. 

3.  A  Parallelogram  is  a  quadrilateral  having  two  pairs 
of  parallel  sides. 


Trapezium  „  Trapezoid.          Rhomboid.          Rhombus.  Rectangle, 


403.     Parallelograms  are  of  three  classes: 

1.  A  Rhomboid   is  a  parallelogram  having   one  pair  of 
parallel  sides  greater  than  the  other,  and  no  right  angle. 

2.  A  Rhombus  is  a  parallelogram    whose  four  sides  are 
equal. 

3.  A  Rectangle  is  a  parallelogram  whose  angles  are  all 
right  angles ;  when  the  rectangle  has  four  equal  sides,  it  is 
a  square. 

A  square  is  a  rhomhns  whose  angles  are  90° ;  it  is  also  the 
form  of  the  unit  for  surface  measure.  It  may  properly  be 
denned,  an  equilateral  rectangle. 


Square. 


382  RAY'S  HIGHER  ARITHMETIC. 

AKEAS. 
TRIANGLES  AND  QUADRILATERALS. 

404.  The  general  rules  depend  on  the  principles  stated 
in  the  following  remarks: 

EEMAKKS. — 1.  The  area  of  a  rectangle  is  equal  to  the  product 
of  its  length  by  its  breadth.  (Art.  197,  Ex.) 

2.  The  diagonal  of  a  rectangle  divides  it  into  two  equal  triangles. 
The   accompanying   figure   illustrates  this; 

EHI  =  HEF.  Observe  also  that  the  per- 
pendicular GL  divides  the  whole  into  two 
rectangles;  EGL  is  half  of  one  of  them, 
LGI  the  half  of  the  other,  and  both  these 
smaller  triangles  make  EGI,  which  must 
therefore  be  half  of  the  whole;  EGI  and 
EHI  have  the  same  base  and  equal  altitudes. 

If  the  triangle  EGH  be  supposed  to  stand  on  GH  as  a  base, 
its  altitude  is  EF;  the  perpendicular  which  represents  the  height 
is,  in  such  a  case,  said  to  fall  on  the  base  produced;  i.  e.,  extended. 
The  triangle  EGH  is  equal  to  the  half  of  GHIL.  Any  triangle 
has  an  area  equal  to  the  product  of  half  the  base  by  the  altitude. 

Observe  also  that  the  trapezoid  EFGI  is  made  of  two  triangles, 
EGF  and  EGI;  each  of  these  has  the  altitude  of  the  trapezoid, 
and  each  has  one  of  the  parallel  sides  for  a  base.  Hence,  the  area 
of  each  being  J  its  base  X  the  common  altitude,  the  two  areas,  or 
the  whole  trapezoid,  must  equal  the  half  of  both  bases  X  the  altitude. 

3.  If  the  piece  EGF  were  taken  off  and  put  on  the  right  of  EGHI, 
the  line  EF  being  placed  on  HI,  the  whole  area  would  be  the  same,  but 
the  perimeter  would  be  inweased,  and  the  figure  would  be  a  rhomboid. 
Different  quadrilaterals  may  have  equal  areas  and  unequal  boundaries; 
also,  they  may  have  the  sides  in  the  same  order  and  equal,  with  unequal 
areas.     To  find  accurately  the  area  of  a  quadrilateral,  more  must 
be  known  than  merely  the  four  sides  in  order.     A  regular  polygon 
has  a  greater  area  than  any  other  figure  of  the  same  perimeter. 

4.  When  triangles  have  equal  bases  their  areas  are  to  each  other 
as  their  altitudes;  the  altitudes  being  equal,  their  areas  are  as  their 
bases.     The  area  of   any  triangle  is  equal   to  half   the  product  of 
the  perimeter  by  the  radius  of  the  inscribed  circle. 


MENSURATION.  383 


GENERAL  RULES. 

1.  To  find  the  area  of  a  parallelogram. 

Rule. — Multiply  one  of  two  parallel  sides  by  the  perpendicular 
distance  between  them. 

II.  To  find  the  area  of  a  triangle. 

Rule. — Take  half  the  product  of  the  base  by  the  altitude. 

III.  To  find  the  area  of  a  trapezoid. 

Rule. — Multiply  half  the  sum  of  the  parallel  sides  by  the 
altitude. 

NOTE. — The  following  is  demonstrated  in  Geometry : 

IV.  To  find  the  area  of  a  triangle  when  the  sides 
are  given.  % 

Rule. — Add  the  three  sides  together  and  take  half  the  sum ; 
from  the  half  sum  take  the  sides  separately ;  multiply  the  half  sum 
and  the  three  remainders  together,  and  extract  the  square  root  of 
the  product. 

NOTES. — 1.  The  area  of  a  trapezium  may  be  found  by  applying 
this  rule  to  the  parts  when  the  sides  are  known  and  the  diagonal  is 
given  in  length  and  in  special  position  as  between  the  sides.  The 
area  of  any  polygon  may  be  found  by  dividing  it  into  triangles  and 
measuring  their  bases  and  altitudes. 

2.  The  area  of  a  rhombus  is  equal  to  half  the  product  of  its  diag- 
onals ;  these  are  at  right  angles. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  area  of  a  parallelogram  whose  base  is  9  ft.  4 
in.  and  altitude  2  ft.  5  in.  22  sq.  ft.  80  sq.  in. 

2.  Of  an  oil  cloth  42  ft.  by  5  ft.  8  in.  26|  sq.  yd. 


384  RAY'S  HIGHER  ARITHMETIC. 

3.  How  many  tiles  8  in.  square  in  a  floor  48  ft.  by  10  ft.? 

1080. 

4.  Find  the  area  of  a  triangle  whose  base  is  72  rcl.  and 
altitude  16  rd.  3  A.  96  sq.  rd. 

5.  Base  13  ft.  3  in.;  altitude  9  ft.  6  in. 

62  sq.  ft.  135  sq.  in. 

6.  Sides  1  ft.  10  in.;  2  ft;  3  ft.  2  in. 

1  sq.  ft.  102— sq.  in. 

7.  Sides  15  rd. ;  18  rd. ;  25  rd.  133.66-  sq.' rd. 

8.  What  is  the  area  of  a  trapezoid  whose  bases  are  9  ft. 
and  21  ft.,  and  altitude  16  ft.?  240  sq.  ft. 

9.  Bases  43  rd.  and  65  rd. ;   altitude  27  rd.  ? 

9  A.  18  sq.  rd. 

10.  What  is  the  area  of  a  figure  made  up  of  3  triangles 
whose  bases  are  10,  12,  16  rd.  and  altitudes  9,  15,  10^  rd.  ? 

1  A.  59  sq.  rcl. 

11.  Whose  sides  are  10,  12,  14,  16  rd.  in  order,  and  dis- 
tance from  the  starting  point  to  the  opposite  corner,  18  rd.  ? 

1  A.  3.9—  sq.  rd. 

12.  How  much  wainscoting  in  a  room  25  ft.  long,  18  ft. 
wide,  and  14  ft.  3  in.  high,  allowing  a  door  7  ft.  2  in.  by  3 
ft.  4  in.,  and  two  windows,  each  5  ft.  8  in.  by  3  ft.  6  in., 
and  a  chimney  6  ft.  4  in.  by  5  ft.  6  in. ;  charging  for  the 
door  and  windows  half- work?  128-|f  sq.  yd. 

13.  What  is  the  perimeter  of  a  rhombus,  one  diagonal  be- 
ing 10  rd.,  and  the  area  86.60J  sq.  rd.?  40—  rd. 

14.  Find   the  cost  of  flooring  and  joisting  a  house  of  3 
floors,  each  48  ft.  by  27  ft.,  deducting  from  each  floor  for  a 
stairway  12  ft.  by  8  ft.  3  in.,  allowing  9  in.  rests  for  the 
joists;  estimating  the  flooring  and  joisting  between  the  walls 
at  $1.46  a  sq.  yd.,  and  the  joisting  in  the  walls  at  76  ct.  a 
sq.  yd. ;  each  row  of  rests  being  measured  48  ft.  long  by  9 
in.  wide.  $600.78 

15.  What  is  the  area  of  a  square  farm  whose  diagonal  is 
20.71  ch.  longer  than  a  side?  250  acres. 

16.  How  many  sq.   yd.   of  plastering  in  a  room  30   ft. 


MENSURATION.  385 

long,  25  ft.  wide,  and  12  ft.  high,  deducting  3  windows, 
each  8  ft.  2  in.  by  5  ft. ;  2  doors  each  7  ft.  by  3  ft.  6  in. ; 
and  a  fire-place  4  ft.  6  in.  by  4  ft.  10  in. ;  the  sides  of  the 
windows  being  plastered  15  in.  deep?  And  what  will  it 
cost,  at  25  ct,  a  sq.  yd.?  215£  sq.  yd.;  cost  $53.83 

17.  From  a  point  in  the  side  and  8  ch.  from  the  corner 
of  a  square  field  containing  40  A.,  a  line  is  run,  cutting  off' 
19J  A.:  how  long  is  the  line?  One  answer,  20|  ch. 

18.  How  much  painting  on  the  sides  of  a  room  20  ft. 
long,  14  ft.   6  in.  wide,  and  10  ft.  4  in.  high,  deducting  a 
fire-place  4  ft.  4  in.  by  4  ft.,  and  2  windows  each  6  ft.  by 
3  ft.  2  in.  ?  73^7  sq.  yd. 

19.  Find  the  cost  of  glazing  the  windows  of  a  house  of  3 
stories,  at  20  ct.  a  sq.  ft.     Each  story  has  4  windows,  3  ft. 
10  in.   wide;    those  in  the  1st  story  are  7  ft.   8  in.  high; 
those  in  the  2nd,  6  ft.  10  in.  high;  in  the  3d,  5  fU  3  in. 
high.  $60.56§ 

REGULAR  POLYGONS  AND  THE  CIRCLE. 

405.  Any  regular  polygon   may  be  divided  into  equal 
isosceles  triangles,  by  lines  from  the  center  to  the  vertices; 
the  apothem  is  their  common  altitude,  and  the  perimeter 
the  sum  of  their  bases. 

406.  To  find  the  area  of  a  regular  polygon. 
Rule. — Multiply  the  perimeter  by  half  the  apothem. 

All   regular  polygons  of   the  same   number  of  sides   are  similar 
figures.     (Art.  389,  Rein.  1.) 

407.  1.  The  circle,  as  already  defined  (Art.  204),  is  a 
figure  bounded  by  a  uniform  curve. 

2.  Any  line  drawn  in  a  circle,   having  its  ends   in   the 

curve,  is  called  a  chord;  as  AB,  BD. 

IT.  A.  r,. 


386 


RAY'S  HIGHER  ARITHMETIC. 


3.  The   portion   of  the   curve  which 
is  cut  off  by  such  a  line  is  called  an 
arc,  and  the  space  between  the  chord 
and  the  arc  is  called  a  segment. 

Thus,  the  curve  APB  is  an  arc,  AB  is 
the  chord  of  that  arc,  and  these  inclose  a 
segment  whose  base  is  AB,  and  whose  height 
is  OP. 

4.  If  a  line  be  drawn  from  the  middle  of  a  chord  to  the 
center,  it  will  be  perpendicular  to  the  chord;  so  also,  a  line 
perpendicular  at  the  middle  of  a  chord,  will,  if  extended, 
pass  through  the  center,  and  bisect  either  of  the  arcs  stand- 
ing on  that  chord.      Thus,  AB   is  bisected  by  the  perpen- 
dicular CO,  arid  the  arc  AP^PB;  so  the  arc  AD  =  BD. 

5.  A  tangent  to  a  circle  as  a  straight  line   having  only 
one  point  in   common   with   the   curve;    it  simply   touches 
the  circle;  a  secant  enters  the  figure  from  without. 

If  with  C  as  a  center,  and  CO  as  a  radius,  a  circle  were  drawn 
in  the  equilateral  triangle  ABD,  the  sides  would  be  tangent  to 
the  circle;  the  circle  would  be  inscribed  in  the  triangle.  The  circle 
of  which  CB  is  radius,  is  circumscribed  about  ABD. 

6.  The  space  inclosed  by  two  radii  and  an  arc,  is  called 
a  sector;  as,  ACP. 

The  arc  of  that  sector  is  the  same  fraction  of  the  whole  cir- 
cumference that  the  area  of  the  sector  is  of  the  whole  circle. 


CALCULATIONS  PERTAINING  TO  THE  CIRCLE. 

408.  The  accompanying  diagrams  present  (Fig.  1)  a 
regular  polygon  of  six  sides,  (Fig.  2)  one  of  twelve  sides, 
and  (Fig.  3)  a  circle  divided  into  twenty-four  sectors. 

REMARKS. — 1.  The  hexagon  is  composed  of  six  equilateral  triangles, 
and  hence  if  OB  be  1,  the  side  AB  —  1,  and  it  is  easy  to  compute  the 
apothem,  V  1  —  \  ~  .86602540378 


MEN8UEA  WON. 


387 


2.  If  the  distance   from  center  to  vertices  be  unchanged,  and  a 
regular  polygon  of  twelve  sides  be  formed  about  the  same  center, 
it  will  differ  less  from  a  circle  whose   ra- 
dius is  OB,  than  the  hexagon  differs  from  /' 

such  a  circle.  This  is  evident  from  the 
second  figure;  and  if  the  polygon  be  made 
of  twenty-four  sides  (the  distance  from  cen- 
ter to  vertices  remaining  the  same),  it  will 
be  still  nearer  the  circle  in  shape  and  size ; 
in  the  space  of  the  diagram,  one  of  the 
twenty-four  triangles  forming  such  a  poly- 
gon would  differ  very  little  from  one  of 
the  twenty-four  sectors  here  shown.  The 
circle  is  regarded  as  composed  of  an  in- 
finite number  of  triangles  whose  common 
altitude  is  the  radius  and  the  sum  of  whose 
bases  is  the  circumference.  Hence,  the  area 
=  J  the  sum  of  bases  X  altitude;  or, 

Area  of  circle  -—  J  circumference  X  radius. 
Area  of  circle  =  }  circumference  X  diameter. 

3.  Since  the  perimeter  of  the  hexagon  is 
6,  it  is  easy  to  compute  the  next  perime- 
ter shown,    which  is  12  times  AP  or    BP. 
The   apothem  being  found  above,  subtract 
it  from  OP  or   1,  and  obtain  .13397459622 
the    perpendicular    of    a   right-angled    tri- 
angle;  then,  the  base  of  that   triangle  be- 
ing .5,  the  half    of  AB,  find  the  hypothe- 
nuse  .517638090205,  =  PB.       Now,   if  we 

treat   PB   as    we   treated   AB  we   can  find  Fig.  3. 

the  apothem  of  the  second  figure,  and  then 

find  one  of  the  24  sides  of  another  polygon,  still  more  nearly  equal  to 

the  circle.      If  these  operations  be   continued,  we  shall  find  results 

8th  and  9th  as  follows: 

Perimeter  of  polygon  of  1536  sides  —  6.28318092 
Perimeter  of  polygon  of  3072  sides  =  6.28318420 

If  the  distance  from  center  to  vertices   be   taken   i   instead  of   1, 
the  results  will  be 

3.141590+    and   3.141592  + 


388  RAY'S  HIGHER  ARITHMETIC. 

Hence,  if  the  circle  of  diameter  1,  be  taken  as  a  polygon  of  1536 
sides,  and  then  as  a  polygon  of  3072  sides,  the  expressions  for 
perimeter  do  not  differ  at  the  fourth  decimal  place.  The  number 
3.1416  is  usually  given,  although  by  more  expeditious  methods  than 
that  above  illustrated,  the  calculation  has  been  carried  to  a  great 
number  of  decimal  places,  of  which  the  following  correctly  shows 
eighteen  : 

3.141592653589793238 

This  important  ratio,  of  circumference  to  diameter,  is  represented 
by  the  Greek  letter  TT  (pi.}. 

4.  Since  circumference  =  diameter  X  ^  and  area  =  \  circum- 

ference X  diameter,  we  have  area  =  j-  X  square  of  diameter.     Rep- 

resenting the  circumference  by  c,  area  by  CJ  diameter  by  (/,  radius 
by  R,  we  have  the  following  formulas: 


W 


GENERAL  RULES. 

409.     Pertaining    to    the    circle    we    have    the    following 
general  rules: 

I.  To  find  the  circumference: 

1.  Multiply  the  diameter  by  3.1415926;  or, 

2.  Divide  the  area  by  J  of  the  diameter;  or, 

3.  Extract  the  square  root  of  12.56637  times  the  area. 

II.  To  find  the  diameter: 

1.  Divide  the  circumference  by  3.1415926;  or, 

2.  Divide  the  area  by  .785398,  and  extract  the  square  root. 

III.  To  find  the  area: 

1.  Multiply  the  diameter  by  J  of  the  circumference;  or, 

2.  Multiply  the  square  of  the  diameter  by  .785398;  or, 

3.  Multiply  the  square  of  the  radius  by  3.1415926 


MENSURA  TION.  389 

IV.  To  find  the  area  of  a  sector  of  a  circle : 

1.  Multiply  the  arc  by  one  half  the  radius;  or, 

2.  Take  such  a  fraction  of  the  whole  area  as  the  arc  is  of  the 
whole  circumference. 

V.  To  find  the  area  of  a  segment  less  than  a  semi- 
circle : 

1.  Subtract  from  the  area  of  the  sector  having  the  same  arc, 
the  area  of  the  triangle  whose  base  is  the  base  of  the  segment,  and 
whose  vertex  is  the  center  of  the  circle;  or, 

2.  Divide  the  cube  of  the  height  by  twice  the  base,  and  increase 
the  quotient  by  two  thirds  of  the  product  of  height  and  base. 

REMARK. — Add  the  triangle  to  the  sector,  if  the  segment  be  greater 
than  a  semicircle.     The  second  rule  gives  an  approximate  result. 

NOTES. — 1.  The  side  of  a  square  inscribed  in  a  circle  is  to  radius 
as  1/2  is  to  1. 

2.  The  side  of  an  inscribed  equilateral  triangle  is  to  radius  as  V  3 
is  to  1. 

3.  If  radius  be  1,  the  side  of  the  inscribed  regular  pentagon  is 
1.1755;  heptagon,  .8677;  nonagon,  .6840;  undecagon,  .5634 

EXAMPLES  FOR  PRACTICE. 

1.  What  are  the  circumferences  whose  diameters  are  16, 
22i,  72.16,  and  452  yd.? 

50.265482;  69.900436;  226.6973;  1420  yd. 

2.  What  are  the  diameters  whose  circumferences  are  56, 
1821,  316.24,  and  639  ft.? 

17.82539;  58.09;  100.66232;  and  203.4  ft. 

3.  Find  the  areas  of  the  circles  with   diameters   10   ft. ; 
2  ft.  5  in.  ;   13  yd.  1  ft. 

78.54  sq.  ft.  ;  660.52  sq.  in.  ;   139  sq.  yd.  5.637  sq.  ft. 

4.  Whose  circumferences  are  46   ft. ;  7  ft.  3   in. ;  6  yd. 
1  ft.  4  in. 

168.386  sq.  ft. ;  4  sq.  ft.  26.322  sq.  in. ;  29.7443  sq.  ft 


390 


RA  Y'S  HIGHER  ARITHMETIC. 


5.  Circum.  47.124  ft.,  diameter  15  ft.          176.715  sq.  ft. 

6.  If  we  saw  down  through  \  of  the  diameter  of  a  round 
log  uniformly  thick,  what  portion  of  the  log  is  cut  in  two? 

.2918 

7.  What  fraction  of  a  round  log  of  uniform  thickness  is 
the  largest  squared  stick  which  can  be  cut  out  of  it? 

.6366 

SOLIDS. 

DEFINITIONS. 

410.  1.  A  Solid  is  that  which  has  length,  breadth,  and 
thickness. 

A  solid  may  have  plane  surfaces,  curved  surfaces,  or  both.  A 
cumed  surface  is  one  no  part  of  which  is  ji  plane. 

2.  The  faces  of  a  solid  are  the  polygons  formed   by  the 
intersections  of  its  bounding  planes ;  the  lines  of  those  inter- 
sections are  called  edges. 

3.  A  Prism  is  a  solid  having  two  bases  which  are  parallel 
polygons,  and  faces  which  are  parallelograms. 

A  prism  is  triangular,  quadrangular,  etc.,  according  to  the  shape 
of  its  base.  The  first  of  the  figures  here  given  represents  a  quadran- 
gular prism,  the  second  a  pentagonal  prism. 


4.  A  right  prism  is  a  prism  whose  faces  are  rectangles. 

5.  A  Parallelepiped  is  a  prism  whose  faces  are  parallel- 
ograms.    Its  bounding  surfaces  are  six  parallelograms. 

The  first  figure  above  represents  a  parallelepiped  whose  faces  are 
rectangles. 


MENSURATION.  391 

6.  A  Cube  is  a  parallelepiped  whose  faces  are  squares. 

7.  A   Cylinder  is  a  solid  having  two   bases  which  are 
equal  parallel  circles,  and  having  an  equal  diameter  in  any 
parallel  plane  between  them. 


8.  A  Pyramid  is  a  solid  with  only  one  base,  and  whose 
faces  are  triangles  with  a  common  vertex. 

9.  A  Cone  is  a  solid  whose  base  is  a  circle,  and  whose 
other  surface  is  convex,  terminating  above  in  a  point  called 
the  vertex. 

10.  A  frustum  of  a  pyramid  or  cone  is  the  solid  which 
remains  when   a  portion   having  the  vertex  is  cut  off  by  a 
plane  parallel  to  the  base. 

11.  A  Sphere   is  a  solid  bounded  by 
a  curved  surface,   every  point  of  which 
is    at    the   same   distance    from    a   point 
within,    called  the  center.     The  diameter 
of  a    sphere   is    a  straight    line    passing 
through  the  center  and  having  its  ends 
in  the  surface;  the  radius  is  the  distance 
from  the  center  to  the  surface. 

A  segment,  of  a  sphere  is  a  portion  cut  off  by  one  plane,  or  between 
two  planes;  its  bases  are  cirdes,  and  its  height  is  the  portion  of  the 
diameter  which  is  cut  off  with  it. 

12.  The   slant  height  of  a  pyramid  is  the  perpendicular 
distance  from  the  vertex  to  one  of  the  sides  of  a  base ;  the 
slant  height  of  a  cone  is  the  straight  line  drawn  from  the 
vertex  to  the  circumference  of  the  base. 


392  RAY'S  HIGHER  ARITHMETIC. 

13.  The  altitude  of  any  solid  is  the  perpendicular  dis- 
tance between   the  planes  of  its  bases,  or  the  perpendicular 
distance  from  its  highest  point  to  the  plane  of  the  base. 

14.  The  Volume  of  a  solid  is  the  number  of  solid  units 
it  contains ;  the  assumed  unit  of  measure  is  a  cube.     (Art. 
199.) 

15.  Solids  are   similar   when  their  like  lines  are  propor- 
tional, and  their  corresponding  angles  equal. 


GENERAL  RULES. 

I.  To   find    the  convex    surface  of  a  prism  or  cyl- 
inder : 

Rule. — Multiply  the  perimeter  of  the  base  by  the  altitude. 

II.  To  find  the  volume  of  a  prism  or  cylinder : 
Rule. — Multiply  the  base  by  the  altitude. 

III.  To   find  the   convex  surface  of  a    pyramid    or 
cone: 

Rule. — Multiply  the  perimeter  of  the  base  by  one  half  the  slant 
height. 

IV.  To  find  the  volume  of  a  pyramid  or  cone: 
Rule. — Multiply  the  base  by  one  third  of  the  altitude. 

V.  To   find    the  convex  surface   of  a   frustum  of  a 
pyramid  or  cone  :  • 

"Rale.— Multiply  half  the  sum  of  the  perimeters  of  the  bases 
by  the  slant  height. 

VI.  To  find  the  solidity  of  a  frustum  of  a  pyramid 
or  cone: 


MENSURATION.  393 

Rule. —  To  the  sum  of  the  two  bases  add  the  square  root  of 
their  product,  and  multiply  the  amount  by  one  third  of  the  alti- 
tude. 

VII.  To  find  the  surface  of  a  sphere : 
Rule. — Multiply  tJie  circumference  by  the  diameter. 

VIII.  To  find  the  volume  of  a  sphere: 
Rule. — Multiply  Hie  cube  of  the  diameter  by  .5235987 

NOTES. — 1.  Similar  solids  are  to  each  other  as  the  cubes  of  their 
like  dimensions. 

2  The  sphere  is  regarded  as  composed  of  an  infinite  number  of 
cones  whose  common  altitude  is  the  radius,  and  the  sniii  of  whose 
bases  is  the  whole  surface  of  the  sphere. 

3.  The  cone  is  regarded  as  a  pyramid  of  an  infinite  number  of  faces, 
and  the  cylinder  as  a  prism  of  an  infinite  number  of  faces. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  convex  surface  of  a  right  prism  with  altitude 
11^  in.,  and  sides  of  base  5^,  6^,  8J,  10^,  9  in. 

450  sq.  in. 

2.  Of  a  right  cylinder  whose  altitude  is  If  ft.,  and  the 
diameter  of  whose  base  is  1  ft.  2^  in. 

6  sq.  ft.,  92.6  sq.  in. 

3.  Find  the  whole  surface  of  a  right  triangular  prism,  the 
sides  of  the  base  60,  80,  and  100  ft.  ;  altitude  90  ft. 

26400  sq.  ft. 

4.  The  whole  surface  of  a  cylinder ;  altitude  28  ft.  ;  cir- 
cumference of  the  base  19  ft.  589.455  sq.  ft. 

5.  Find  the  convex  surface  and  whole  surface  of  a  right 
pyramid   whose   slant   height   is  391   ft.  ;  the   base   640   ft. 
square. 

Conv.  surf.  500480  sq.  ft. ;  whole  surf.   910080  sq.  ft. 


394  KAY'S  HIGHER  ARITHMETIC. 

6.  Of  a   right   cone  whose  slant  height  is   66  ft.  8  in. ; 
radius  of  the  base  4  ft.  2  in. 

125663.706  sq.  in. ;  133517.6876  sq.  in. 

7.  Find  the  solidity  of  a  pyramid  whose  altitude  is  1  ft. 
2  in.,  and  whose  base  is  a  square  4J  in.  to  a  side. 

94^  cu.  in. 

8.  Whose  altitude  is  15.24  in.,  and  whose  base  is  a  triangle 
having  each  side  1  ft.  316.76  cu.  in. 

9.  What  is  the  solidity  of  a  prism  whose  bases  are  squares 
9  in.  on  a  side,  and  whose  altitude  is  1  ft.  7  in.  ? 

1539  cu.  in. 

10.  Whose  altitude  is  6^  ft. ,  and  whose  bases  are  parallel- 
ograms 2  ft.  10  in.  long  by  1  ft.  8  in.  wide? 

30  cu.  ft.  1200  cu.  in. 

11.  Whose  altitude  is  7  in.,  and  whose  base  is  a  triangle 
with  a  base  of  8  in.  and  an  altitude  of  1  ft.  ?         336  cu.  in. 

12.  Whose  altitude  is  4  ft.  4  in.,  and  whose  base  is  a  tri- 
angle with  sides  of  2,  2J,  and  3  ft.  ?  10.75  cu.  ft. 

13.  What  is  the  solidity  of  a  cylinder  whose  altitude  is  10^ 
in.,  and  the  diameter  of  whose  base  is  5  in.  ?     206.167  cu.  in. 

14.  Find  the  convex  surface  of  a  frustum  of  a  pyramid 
with  slant  height  3J  in.,  lower  base  4  in.  square,  upper  base 
2|  in.  square.  43J  sq.  in. 

15.  The  convex  surface  and  whole  surface  of  the  frustum 
of  a  cone,  the  diameters  of  the  bases  being  7  in.  and  3  in., 
and  the  slant  height  5  in. 

Conv.  surf.  78.5398  sq.  in,,  whole  surf.  124.0929  sq.  in. 

16.  Find   the  solidity  of  a  frustum  of  a  pyramid  whose 
altitude  is  1  ft.  4|  in.  ;  lower  base,  lOf  in.  square ;  upper, 
4^  in.  square.  974^f|  cu.  in. 

17.  Of  a  frustum  of  a  cone,  the  diameters  of  the  bases 
being  18  in.  and  10  in.,  and  the  altitude  16  in. 

2530.03  cu.  in. 

18.  What  are  the  surfaces  of  two  spheres  whose  diameters 
are  27  ft.  and  10  in.  ? 

2290.221  -f  sq.  ft.  and  314.16  sq.  in. 


MENSURATION.  395 

19.  Find  the  solidity  of  a  sphere  whose  diameter  is  6  mi., 
and  surface  113.097335  sq.  mi.  113.097335  cu.  mi. 

20.  Of  a  sphere  whose  diameter  is  4  ft.         33.5103  cu.  ft. 

21.  Of  a  sphere  whose  surface  is  40115  sq.  mi. 

755499J  cu.  mi. 

22.  By  what  must  the  diameter  of  a  sphere  be  multiplied 
to  make  the  edge  of  the  largest  cube  which  can  be  cut  out 
of  it?  .57735 


MISCELLANEOUS  MEASUEEMENTS. 
MASONS'  AND  BRICKLAYERS'  WORK. 

411.  Masons'    work    is    sometimes    measured    by    the 
cubic  foot,  and  sometimes  by  the  perch.     The  latter  is  16^- 
ft.  long,  11  ft.  wide,  and  1  ft.  deep,  and   contains  16^-  X 
1-i-  X  1  =  24|  cu.  ft.,  or  25  cu.  ft.  nearly. 

412.  To   find  the   number  of  perches  in  a  piece  of 
masonry. 

Rule. — Find  the  solidity  of  the  wall  in  cubic  feet  by  the  rules 
given  for  mensuration  of  solids,  and  divide  it  by  24f . 

NOTE. — Brick  work  is  generally  estimated  by  the  thousand  bricks ; 
the  usual  size  being  8  in.  long,  4  in.  wide,  and  2  in.  thick.  When 
bricks  are  laid  in  mortar,  an  allowance  of  T^  is  made  for  the  mortar. 

EXAMPLES  FOR  PRACTICE. 

1.  How  many  perches  of  25  cu.  ft.  in  a  pile  of  building- 
stone  18  ft.  long,  8J  ft.  wide,  and  6  ft.  2  in.  high  ? 

37.74  =  37f  perches  nearly. 

2.  Find  the  cost  of  laying  a  wall  20  ft.  long,  7  ft.  9  in. 
high,  and  with  a  mean  breadth  of  2  ft.,  at  75  ct.  a  perch. 

«9.39 


396  HAY'S  HIGHER  ARITHMETIC. 

3.  The  cost  of  a  foundation  wall  1  ft.  10  in.  thick,  and 
9  ft:  4  in.  high,  for  a  building  36  ft.  long,  22  ft.  5  in.  wide 
outside,  at  $2.75  a  perch,  allowing  for  2  doors  4  ft.  wide. 

$192.98 

4.  The  cost  of  a  brick  wall  150  ft.  long,  8  ft.  6  in.  high, 
1  ft.  4  in.  thick,  at  $7  a  thousand,  allowing  T^  for  mortar  ? 

$289.17 

5.  How  many  bricks  of  ordinary  size  will  build  a  square 
chimney  86  ft.  high,  10  ft.  wide  at  the  bottom,  and  4  ft.  at 
the  top  outside,  and  3  ft.  wide  inside  all  the  way  up? 

89861+  bricks. 

SUGGESTION. — Find  the  solidity  of  the  whole  chimney,  then  of  the 
hollow  part ;  the  difference  will  be  the  solid  part  of  the  chimney. 

GAUGING. 

413.  Gauging  is  finding  the  contents  of  vessels,  in  bush- 
els, gallons,  or  barrels. 

414.  To  gauge  any  vessel  in  the  form   of  a  rect- 
angular solid,  cylinder,  cone,  frustum  of  a  cone,  etc. 

Rule. — Find  the  solidity  of  the  vessel  in  cubic  inches  by  the 
rules  already  given;  this  divided  by  2150.42,  will  give  the  con- 
tents in  bushels;  by  231,  will  give  it  in  wine  gallons,  which 
may  be  reduced  to  barrels  by  dividing  the  uuniber  by  31^. 

NOTE. — In  applying  the  rule  to  cylinders,  cones,  and  frustums  of 
cones,  instead  of  multiplying  the  square  of  half  the  diameter  by 
3.14159265,  and  dividing  it  by  231,  multiply  the  square  of  tlw  diameter  by 
.0034,  which  amounts  to  the  same,  and  is  shorter. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  bushels  in  a  bin  8  ft.  3  in.  long,  3  ft.  5  in. 
high,  and  2  ft.  10  in.  wide?  64.18  bu. 


MENSURATION.  397 

2.  How  many  wine  gallons  in  a  bucket  in  the  form  of  a 
frustum  of  a  cone,  the  diameters  at  the  top  and  bottom  being 
13  in  and  10  in.,  and  depth  1%  in.?  5.4264  gal. 

3.  How  many  barrels  in  a  cylindrical  cistern  11  ft.  6  in. 
deep  and  7  ft.  8  in  wide?  126.0733  bbl. 

4.  In  a  vat  in  the  form  of  a  frustum  of  a  pyramid,  5  ft. 
deep,  10  ft.  square  at  top,  9  ft.  square  at  bottom? 

107.26  bbl. 

415.  To  find  the  contents  in  gallons  of  a  cask  or 
barrel. 

REMARK. — When  the  staves  are  straight  from  the  bung  to  each 
end,  consider  the  cask  as  two  frustums  of  a  cone,  and  calculate  its 
contents  by  the  last  rule ;  but  when  the  staves  are  curved,  use  this 
rule: 

Rule. — Add  to  the  head  diameter  (inside)  two  thirds  of  tlie 
difference  between  the  head  and  bung  diameters ;  but  if  the  staves 
are  only  slightly  curved ,  add  six  tenths  of  this  difference;  this  gives 
the  mean  diameter ;  express  it  in  inches,  square  it,  multiply  it  by 
the  length  in  inches,  an$  this  product  by  .0034:  the  product  will 
be  the  contents  in  wine  gallons. 

NOTE. — After  finding  the  mean  diameter,  the  contents  are  found  as 
if  the  cask  were  a  cylinder. 

EXAMPLES  FOR  PRACTICE. 

1.  Find  the  number  of  gallons  in  a  cask  of  beer  whose 
staves  are  straight  from  bung  to  head,  the  length  being  26 
in.,  the  bung  diameter  16  inches,  and  head  diameter  13  in. 

18.65  gal. 

2.  In   a  barrel   of  whisky,  with  staves   slightly   curved, 
length  2  ft.  10  in.,  bung  diameter  1  ft.  9  in.,  head  1  ft.  6  in. 

45.32  gal. 

3.  In  a  cask  of  wine  with  curved  staves,  length  5  ft.  4  in., 
bung  diameter  3  ft.  6  in.,  head  diameter  3  ft.        348.16  gal. 


398  RAY'S  HIGHER  ARITHMETIC. 


LUMBER  MEASURE. 

416.  To  find  the  amount  of  square-edged  inch 
boards  that  can  be  sawed  from  a  round  log. 

REMARK. — The  following  is  much  used  by  lumber-men,  and  is 
sufficiently  accurate  for  practical  purposes.  It  is  known  as  Doyle's 
Kule. 

Rule. — From  the  diameter  in  inches  subtract  4;  the  square, 
of  the  remainder  wilt  be  the  number  of  square  feet  of  inch  boards 
yielded  by  a  log  16  feet  in  length. 


EXAMPLES  FOR  PRACTICE. 

1.  How  much  square-edged  inch  lumber  can  be  cut  from 
a  log  32  inches  in  diameter,  and  20  feet  long? 

OPERATION. 

32  —  4  =  28;   28  X  28  X  ft  =  080  feet. 
Or,  f  X  28  X  28  =  980  feet. 

2.  In  a  log  24  in.  in  diameter,  and  12  ft.  long?    300  ft. 

3.  In  a  log  25  in.  in  diameter,  and  24  ft.  long? 

661i  ft. 

4.  In  a  log  50  in.  in  diameter,  and  12  ft.  long? 

1587  ft. 

To  MEASURE  GRAIN  AND  HAY. 

417.  Grain  is  usually  estimated  by  the  bushel,  and  sold 
"by  weight;  Hay,  by  the  ton. 

REMARKS. — 1.  The  standard  bushel  contains  2150.4  cubic  inches. 
A  cubic  foot  is  nearly  .8  of  a  bushel. 

2.  Hay  well  settled  in  a  mow  may  be  estimated  (approximately) 
at  550  cubic  feet  for  clover,  and  450  cubic  feet  for  timothy,  per  ton. 


MENSURATION.  399 

418.  To  find  the  quantity  of  grain  in  a  wagon  or 
in  a  bin: 

Rule. — Multiply  the  contents  in  cubic  feet  by  .8 

EEMABKS. — 1.  If  it  be  corn  on  the  cob,  deduct  one  half. 

2.  For  corn  not  u  shucked,"  deduct  two  thirds  for  cob  and  shuck. 

419.  To  find  the  quantity  of  hay  in  a  stack,  rick, 
or  mow  : 

Rule. — Divide  the  cubical  contents  in  feet  by  550  for  clover, 
or  by  450  for  timothy;  the  quotient  will  be  the  number  of  tons. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many   bushels  of  shelled   corn,   or  corn  on   the 
cob,  or  corn  not  shucked,  will  a  wagon-bed  hold  that  is  10^ 
feet  long,  3J  feet  wide,  and  2  feet  deep? 

58.8  bti.;  29.4  bti. ;    19.6  bu. 

2.  In  a  bin  40  feet  long,  16  wide,  and  10  feet  high? 

5120  bu. 

3.  A  hay-mow  contains  48000  cubic  feet:  how  many  tons 
of  well  settled  clover  or  timothy  will  it  hold? 

87T3T  tons  clover ;   106-|  tons  timothy. 


Topical  Outline. 
MENSURATION. 

1.  General  Definitions :— Geometry,  Magnitude,  Mensuration. 

f  Parallel. 
Perpendicular. 

I   Straight \    Horizontal. 

2-  Lines \   Broken.  |    Vertical. 

Curved.  Diagonal. 


400 


JRAY'JS  HIGHER   ARITHMETIC. 


MENSURATION. — (Continued. ) 


3.  Angles  . 


f   Right. 
\  Oblique 


4.  Surfaces. 


5.  Solids 


Right. 

ie /  Acute. 

\  Obtuse. 

1.  General  Definitions:— Plane,  Plane   figure,   Area,    Polygons,' 

Regular,  Perimeter,  Similar,  Center,  Altitude,  Base, 
Apothem. 

1.  Right /  IsosceU*' 

I  Sc 

2.  Triangles... 

(Rules.) 

1   2.  Oblique- 
Obtuse , 


Scalene. 
Acute 


{  Scalene, 
•j  Isosceles. 


3.  Quadrilateral.. 
(Rules.) 


4.  Circle. 


Equilateral. 

Scalene. 

Isosceles. 

1.  Trapezium. 

2.  Trapezoid.  ["  L  Rhomboid. 

3.  Parallelogram.  J   2.  Rhombus...  \  Square. 

!   3.  Rectangle.. .j  Square. 

1.  Terms:— Circumference,  Radius,  Chord,  Diam- 

eter, Segment,  Sector,  Tangent,  Secant. 

2.  Calculations,  Value  of  IT. 
I  3.  Formulas,  Rules. 

1.  General  Definitions:— Solid,  Base,  Face,  Edge,  Similar. 


2.  Prism 

(Rules.) 


Triangular.    (Right.) 

Quadrangular < 

j   Pentagonal,  etc. 

c  Altitude. 

3.  Pyramid J  giant  Height. 

(Rules.)    (Frustum.)      (  Convex  Surface. 

{Altitude. 
Slant  Height. 
Convex  Surface. 
f  Surface. 


Parallelepiped.  cCube 
(Right.) 


5.  Cylinder 

(Rules.) 


\  Solidity. 


6.  Sphere.. 


7.  General  Formulas. 

8.  Miscellaneous 

Applications 


(1.  Terms : — Radius,  Diameter,  Seg- 
ment. 
2.  Convex  Surface,  Rule. 
3.  Solidity,  Rule. 

'  Masons'  and  Bricklayers'  Work. 

Gauging. 

Lumber  Measure. 
.  Measuring  Grain  and  Hay. 


XXIII.  MISCELLANEOUS  EXERCISES. 

NOTE.— Nos.  1  to  50  are  to  be  solved  mentally. 

1.  If  I  gain  J  ct.  apiece  by  selling  eggs  at  7  ct.  a  dozen,  how 
much  apiece  will  I  gain  by  selling  them  at  9  ct.  a  dozen?  -f%  ct. 

2.  If  I  gain  \  ct.  apiece  by  selling  apples  at  3  for  a  dime,  how 
much  apiece  would  I  lose  by  selling  them  4  for  a  dime?  \  ct. 

3.  If  I  sell  potatoes  at  37 \  ct.  per  bu.,  my  gain  is  only  \  of  what  it 
would  be,  if  I  charged  45  ct.  per  bu. :  what  did  they  cost  me? 

26£  ct.  per  bu. 

4.  If  I  sell  my  oranges  for  65  ct.,  I  gain  f  ct.  apiece  more  than  if  I 
sold  them  for  50  ct. :  how  many  oranges  have  1?  40  oranges. 

5.  If  I  sell  my  pears  at  5  ct.  a  dozen,  I  lose  16  ct.;  if  I  sell  them  at 
8  ct.  a  dozen,  I  gain  11  ct. :  how  many  pears  have  I,  and  what  did 
they  cost  me?  9  dozen  at  6J  ct.  per  dozen. 

6.  If  I  sell  eggs  at  6  ct.  per  dozen,  I  lose  f-  ct.  apiece;  how  much  per 
dozen  must  I  charge  to  gain  f  ct.  apiece?  22  ct. 

7.  One  eighth  of  a  dime  is  what  part  of  3  ct.?  •£•$. 

8.  If  I  lose  |  of  my  money,  and  spend  f  of  the  remainder,  what 
part  have  I  left?  Jf . 

9.  A's  land  is  ^-  less  in  quantity  than  B's,  but  jV  better  in  quality: 
how  do  their  farms  compare  in  value?  A's  =  T%  of  B's. 

10.  If  f  of  A's  money  equals  f  of  B's,  what  part  of  B's  equals  |  of 
A's?  f. 

11.  I  gave  A  f\  of  my  money,  and  B  j7^  of  the  remainder:  who  got 
the  most,  and  what  part?  B  got  -^  of  it  more  than  A. 

12.  A  is  f  older  than  B,  and  B  f  older  than  C:  how  many  times 
C's  age  is  A's?  2J. 

13.  Two  thirds  of    my  money  equals  f  of  yours;    if   we  put  our 
money  together,  what  part  of  the  whole  will  I  own?  T6r. 

14.  How  many  thirds  in  J?  1|. 

15.  Reduce  f  to  thirds;  f  to  ninths;  and  f  to  a  fraction,  whose  nu- 

o 

merator  shall  be  8.  2  ?  thirds ;  7 A  ninths ; 

'  i  o  i 
loj. 

16.  What  fraction  is  as  much  larger  than  f  as  f  is  less  than  f  ?    -}|. 

17.  After  paying  out  \  and  \  of  my  money,  I  had  left  $8  more  than 

I  had  spent:  what  had  I  at  first?  $80. 

H.  A.  34.  (401) 


402  RAY'S  HIGHER  ARITHMETIC. 

18.  In  12  yr.  I  shall  be  |  of  my  present  age:  how  long  since  was  J  5 
of  my  present  age?  8f  yr. 

19.  Four  times  f  of  a  number  is  12  less  than  the  number;  what  is 
the  number?  108. 

20.  A  man  left  T5T  of  his  property  to  his  wife,  f  of  the  remainder  to 
his  son,  and  the  balance,  $4000,  to  his  daughter,  what  was  the  estate? 

$22000. 

21.  1  sold  an  article  for  \  more  than  it  cost  me,  to  A,  who  sold  it 
for  $6,  which  was  f  less  than  it  cost  him  :  what  did  it  cost  me?         $8. 

22.  A  is  f  older  than  B ;  their  father,  who  is  as  old  as  both  of  them, 
is  50  yr.  of  age:  how  old  are  A  and  B?  A,  27 J  yr. ;  B,  22 £  yr. 

23.  A  pole  was  f  under  water ;  the  water  rose  8  ft.,  and  then  there 
was  as  much  under  water  as  had  been  above  water  before :  how  long 
is  the  pole?  18|  ft. 

24.  A  is  f  as  old  as  B;  if  he  were  4  yr.  older,  he  would  be  T9^  as  old 
as  B;  how  old  is  each?  A,  20  yr.;  B,  26*  yr. 

25.  A's  money  is  $4  more  than  f  of  B's,  and  $5  less  than  f  of  B's: 
how  much  has  each?  A,  $76  •  B;  $108. 

26.  Two  thirds  of  A's  age  is  f  of  B's,  and  A  is  3^  yr.  the  older: 
how  old  is  each?  A,  31  i  yr.;  B,  28  yr. 

27.  If  3  boys  do  a  work  in  7  hr.,  how  long  will  it  take  a  man  who 
works  4-J-  times  as  fast  as  a  boy?  4|  hr. 

28.  If  6  men  can  do  a  work  in  5J  days,  how  much  time  would  be 
saved  by  employing  4  more  men?  2J  days. 

29.  A  man  and  2  boys  do  a  work  in  4  hr. :  how  long  would  it  take 
the  man  alone  if  he  worked  equal  to  3  boys?  6f  hr. 

30.  A  man  and  a  boy  can  mow  a  certain  field  in  8  hr. ;  if  the  boy 
rests  3f  hr.,  it  takes  them  9J  hr. ;  in  what  time  can  each  do  it? 

Man,  13Jhr.;  boy,  20  hr. 

31.  Five  men  were  employed  to  do  a  work ;  two  of  them  failed  to 
come,  by  which  the  work  was  protracted  4J  days:  in  what  time  could 
the  5  have  done  it?  6|  days. 

32.  Three  men  can  do  a  work  in  5  days;  in  what  time  can  2  men 
and  3  boys  do  it,  allowing  4  men  to  work  equal  to  9  boys?  4J  da, 

33.  A  man  and  a  boy  mow  a  10-acre  field ;  how  much  more  does 
the  man  mow  than  the  boy,  if  2  men  work  equal  to  5  boys?          4|-  A. 

34.  Six  men  can  do  a  work  in  4£  days;  after  working  2  days,  how 
many  must  join  them  so  as  to  complete  it  in  3|  da.?  4  men. 

35.  Eight  men  can  do  a  certain  amount  of  work  in  6f  days;  after 
beginning,  how  soon  must  they  be  joined  by  2  more  so  as  to  complete 
it  in  5J  days?  In  2f  days. 


MISCELLANEOUS  EXERCISES.  403 

36.  Seven  men  can  build  a  wall  in  5J  days;  if  10  men  are  employed, 
what  part  of  his  time  can  each  rest,  and  the  work  be  done  in  the  same 
time?  TV 

37.  Nine  men  can  do  a  work  in  8}  days;  how  many  days  may  3 
remain  away,  and  yet  finish  the  work  in  the  same  time  by  bringing 
5   more  with  them?  5/j  days. 

38.  Ten  men  can  dig  a  trench  in  7-J-  days;  if  4  of  them  are  absent 
the  first  2^  days,  how  many  other  men  must  they  then  bring  with 
them  to  complete  the  work  in  the  same  time?  .  2  men. 

39.  At  what  times  between  6  and  7  o'clock  are  the  hour-hand  and 
minute-hand  20  min.  apart?    10}£  min.  after  6,  and  54T6T  min.  after  6. 

40.  At  what  times  between  4  and  5  o'clock  is  the  minute-hand  as 
far  from  8  as  the  hour-hand  is  from  3? 

32T4F  min.  after  4;  and  49T1T  min.  after  4. 

41.  At  what  time  between  5  and  6  o'clock  is  the  minute-hand  mid- 
way between  12  and  the  hour-hand?  when  is  the  hour-hand  midway 
between  4  and  the  minute-hand? 

13^  min-  after  5;  and  36  min.  after  5. 

42.  A,  B,  and  C  dine  on  8  loaves  of  bread ;  A  furnishes  5  loaves ;  B, 
3  loaves;   C  pays  the  others  8d.  for  his  share:   how  must  A  and  B 
divide  the  money?  A  takes  7d.;  B,  Id. 

43.  A  boat  makes  15  mi.  an  hour  down  stream,  and  10  mi.  an  hour 
up  stream:  how  far  can  she  go  and  return  in  9  hr.?  54  mi. 

44.  I  can  pasture  10  horses  or  15  cows  on  my  ground ;  if  I  have  9 
cows,  how  many  horses  can  I  keep?  4  horses. 

45.  A's  money  is  12  <f0  of  B's,  and  16  </0  of  C's;  B  has  $100  more 
than  C:  how  much  has  A?  $48. 

46.  Eight  men  hire  a  coach;  by  getting  6  more  passengers,  the  ex- 
pense to  each  is  diminished  $1|:  what  do  they  pay  for  the  coach? 

$32}. 

47.  A  company  engage  a  supper;  being  joined  by  f  as  many  more, 
the  bill  of  each  is  60  ct.  less:  what  would  each  have  paid  if  none  had 
joined  them?  $2.10 

48.  By  mixing  5  Ib.  of  good  sugar  with  3  Ib.  worth  4  ct.  a  Ib.  less, 
the  mixture  is  worth  8|  ct.  a  Ib. :  find  the  prices  of  the  ingredients. 

10  ct.  and  6  ct.  a  Ib. 

49.  By  mixing  1.0  Ib.  of  good  sugar  with  6  Ib.  worth  only  f  as 
much,  the  mixture  is  worth  1  ct.  a  Ib.  less  than  the  good  sugar:  find 
the  prices  of  the  ingredients  and  of  the  mixture. 

Ingredients,  8  ct.  and  5^  ct. ;  Mixt.,  7  ct.  per  Ib. 

50.  A  and  B  have  the  same  amount  of  money ;  if  A  had  $20  more, 


404  RAY'S  HIGHER  ARITHMETIC. 

and  B  $10  less,  A  would  have  2^  times  as  much  as  B:  what  amount 
has  each?  $32J. 


51.  A  and  B  pay  $1.75  for  a  quart  of  varnish,  and  10  ct.  for  the 
bottle;  A  contributes  $1,  B,  the  rest:  they  divide  the  varnish  equally, 
and  A  keeps  the  bottle :  which  owes  the  other}  and  how  much  ? 

B  owes  A  2J  ct. 

52.  How  far  does  a  man  walk  while  planting  a  field  of  corn  285  ft. 
square,  the  rows  being  3  ft.  apart  and  3  ft.  from  the  fences? 

5  mi.  6  rd.  6  ft. 

53.  Land  worth  $1000  an  acre,  is  worth  how  much  a  front  foot  of 
90  ft.  depth;  reserving  j1^  for  streets?  $2.295+. 

54.  I  buy  stocks  at  20  ^  discount,  and  sell  them  at  10  ft  premium : 
what  per  cent,  do  I  gain?  37 J  <fa. 

55.  I  invest,  and  sell  at  a  loss  of  15  ^,;  I  invest  the  proceeds  again, 
and  sell  at  a  gain  of  15  <jc :  do  I  gain  or  lose  on  the  two  speculations, 
and  how  many  per  cent.?  Lose  2\  o/G. 

56.  I  sell  at  8  ft  gain,  invest  the  proceeds,  and  sell  at  an  advance  of 
12J  <fc\  invest  the  proceeds  again,  and  sell  at  4  ^  loss,  and  quit  with 
$1166  40:  what  did  I  start  with?  $1000. 

57.  I  can  insure  my  house  for  $2500,  at  ^  cf0  premium  annually,  or 
permanently  by  paying  down  12  annual  premiums:  which  should  I 
prefer,  and  how  much  will  I  gain  by  it  if  money  is  worth  6  <J0  per 
annum  to  me?  The  latter;  gain,  $113.33^ 

58.  A  owes  B  $1500,  due  in  1  yr.  10  man.     He  pays  him  $300  cash, 
and  a  note  of  6  mon.  for  the  balance:  what  is  the  face  of  the  note, 
allowing  interest  at  6  tfcl  $1080.56 

59.  If  I  charge  12  <fi  per  annum  compound  interest,  payable  quar- 
terly, what  rate  per  annum  is  that?  12jV$ftftfo  ^. 

60.  How  many  square  inches  in  one  face  of  a  cube  which  contains 
2571353  cu.  in.?  18769  sq.  in. 

61.  What  is  the  side  of   a   cube  which   contains  as  many  cubic 
inches,  as  there  are  square  inches  in  its  surface?  6  in. 

62.  The  boundaries  of  a  square  and  circle  are  each  20  ft.*,  which 
is  the  greater,  and  how  much?  Circle;  6.831  sq.  ft.  nearly. 

63.  If  I  pay  $1000  for  a  5  yr.  lease,  and  $200  for  repairs,  how  much 
rent  payable  quarterly  is  that  equal  to,  allowing  10  ft  interest? 

$307.92  a  year. 

64.  What  is  the  value  of  a  widow's  dower  in  property  worth  $3000, 
her  age  being  40,  and  interest  5  %t  $669.50 


MISCELLANEOUS  EXERCISES.  405 

65.  What  principal  must  be  loaned  Jan.  1,  at  9  %,  to  be  re-paid  by 
5  installments  of  $200  each,  payable  on  the  first  day  of  each  of  the 
five  succeeding  months?  $978.10 

66.  After  spending  25  <f0  of  my  money,  and  25  %  of  the  remainder, 
I  had  left  $675:  what  had  I  at  first?  $1200. 

67.  I  had  a  60-day  note  discounted  at  1  <f0  a  month,  and  paid  $4.80 
above  true  interest:  what  was  the  face  of  the  note?  $11112.93 

68.  Invested  $10000 ;  sold  out  at  a  loss  of  20% :  how  much  must  I 
borrow  at  4%,  so  that,  by  investing  all  I  have  at  18%,  I  may  retrieve 
my  loss?  $4000. 

69.  If  \  of  an  inch  of  rain  fall,  how  many  bbl.  will  be  caught  by 
a  cistern  which  drains  a  roof  52  ft.  by  38  ft.  ?  9.776+  bbl. 

70.  A  father  left  $20000  to  be  divided  among  his  4  sons,  aged  6 
years,  8  years,  10  years,  and  12  years  respectively,  so  that  each  share, 
placed  at  4J%  compound  interest,  should  amount  to  the  same  when 
its  possessor  became  of  age  (21  yr.) :  what  were  the  shares? 

$4360.34;  $4761.59;  $5199.78;  $5678.29 

71.  $30000  of  bonds  bearing  7%  interest,  payable  semi-annually, 
and  due  in  20  yr.  are  bought  so  as  to  yield  8%  payable  semi-annually: 
what  is  the  price  ?  $27031.08 

72.  A  man   wishes  to  know  how  many  hogs  at  $9,  sheep  at  $2, 
lambs   at   $1,  and  calves  at  $9  per  head,  can  be  bought  for  $400, 
having,  of  the  four  kinds,  100  animals  in  all.     How  many  different 
answers  can  be  given  ?  288  answers. 

73.  The  stocks  of  3  partners,  A,  B,  and  C,  are  $350,  $220,  and  $250, 
and  their  gains  $112,  $88,  and  $120  respectively;  find  the  time  each 
stock  was  in  trade,  B's  time  being  2  mon.  longer  than  A's. 

A's,  8  mon. ;  B's,  10  mon.;  C's,  12  mon. 

74.  By  discounting  a  note  at  20%  per  annum,  I  get  22 \<J0  per  annum 
interest:  how  long  does  the  note  run?  200  days. 

75.  A  receives  $57.90,  and  B  $29.70,  from  a  joint  speculation :  if  A 
invested  $7.83J  more  than  B,  what  did  each  invest  ? 

A,  $16.08 };  B,  $8.25 

76.  A  borrows  a  sum  of  money  at  6%,  payable  semi-annually,  and 
lends  it  at  12%,  payable  quarterly,  and  clears  $2450.85  a  year:  what 
is  the  sum  ?  $38485.87 

77.  Find  the  sum  whose  true  discount  by  simple  interest  for  4  yr. 
is  $25  more  at  6%  than  at  4%  per  annum.  $449.50 

78.  I   invested  $2700  in  stock    at  25^,  discount,  which  pays  8% 
annual  dividends :  how  much  must  I  invest  in  stock  at  4%  discount 
and  paying  10%  annual  dividends,  to  secure  an  equal  income? 

$2764.80 


406  RAY'S  HIGHER  ARITHMETIC. 

79.  Exchanged  $5200  of  stock  bearing  5$  interest  at  69<&,  for  stock 
bearing  7%  interest  at  92$,  the  interest  on  each  stock  having  been 
just  paid :  what  is  my  cash  gain,  if  money  is  worth  6%  to  me? 

$216.66| 

80.  Bought  goods  on  4  mon.  credit;  after  7  rnon.  I  sell  them  for 
$1500,  2J$  off  for  cash;  my  gain  is  15%,  money  being  worth  6$: 
what  did  I  pay  for  the  goods  ?  $1252.94 

81.  The  9th  term  of  a  geometric  series  is  137781,  and  the  13th 
term  11160261 :  what  is  the  4th  term?  567. 

82.  My  capital  increases  every  year  by  the  same  per  cent. ;  at  the 
end  of  the  3d  year  it  was  $13310 ;  at  the  end  of  the  7th  year  it  was 
$19487.171 :  what  was  my  original  capital,  and  the  rate  of  gain? 

$10000,  and  10%. 

83.  Find  the  length  of  a  minute-hand,  whose  extreme  point  moves 
4  inches  in  3  min.  28  sec.  11.02 —  in. 

84.  Three  men  own  a  grindstone,  2  ft.  8  in.  in  diameter :  how  many 
inches  must  each  grind  off  to  get  an  equal  share,  allowing  6  in.  waste 
for  the  aperture?          1st,  2.822—  in. ;  2d,  3.621+  in. ;  3d,  6.557—  in. 

85.  I  sold  an  article  at  20$  gain;  had  it  cost  me  $300  more,  I 
would  have  lost  20$  :  find  the  cost.  $600. 

86.  A  boat  goes  16J  miles  an  hour  down  stream,  and  10  mi.  an 
hour  up  stream :  if  it  is  22J  hr.  longer  in  coming  up  than  in  going 
down,  how  far  down  did  it  go  ?  585  mi. 

87.  Had  an  article  cost  10$  less,  the  number  of  $  gain   would 
have  been  15  more :  what  was  .the  ft  gain  ?  35$. 

88.  Bought  a  check  on  a  suspended  bank  at  55$  ;  exchanged  it  for 
railroad  bonds  at  60$,  which  bear  7$  interest:  what  rate  of  interest 
do  I  receive  on  the  amount  of  money  invested?  2l^^0' 

89.  Bought  sugar  for  refinery;  6$  is  wasted  in  the  process;  30$ 
becomes  molasses,  which  is  sold    at  40    per  cent,  less  than  the  same 
weight  of  sugar  cost ;  at  what  per  cent  advance  on  the  first  cost  must 
the  clarified  sugar  be  sold,  so  as  to  yield  a  profit  of  14$  on  the  invest- 
ment? 50$. 

90.  There  is  coal  now  on  the  dock,  and  coal  is  running  on  also, 
from  a  shoot,  at  a  uniform  rate.     Six  men  can  clear  the  dock  in  one 
hour,  but  11  men  can  clear  it  in  20  minutes :  how  long  would  it  take 
4  men  ?  5  hr. 

91.  A  distiller  sold  his  whisky,  losing  4$  ;  keeping  $18  of  the  pro- 
ceeds, he  gave  the  remainder  to  an  agent  to  buy  rye,  8$  commission ; 
he  lost  in  all  $32 :  what  was  the  whisky  worth  ?  $300. 

92.  A  clock  gaining  3J  rain,  a  day  was  started  right  at  noon  of 
the  22d  of  February,  1804:  what  was  the  true  time  when   that  clock 


MISCELLANEOUS  EXERCISES.  407 

• 

showed  noon  a  week  afterward ;  and,  if  kept  going,  when  did  it  next 
show  true  time?  35  min.  32.9  sec.  after  11  A.  M. 

True,  Sept.  15th,  8f  min.  past  5  A.  M. 

93.  The  number  of  square  inches  in  one  side  of  a  right-angled  tri- 
angular board  is  144,  and  the  base  is  half  the  height;  required  the 
areas  of  the  different  triangles  which  can  be  marked  off  by  lines  par- 
allel to  the  base,  at  12,  13,  14,  14J  inches  from  the  smaller  end. 

36  sq.  in. ;  42^  sq.  in. ;  49  sq.  in. ;  52T9g  sq.  in. 

94.  Suppose  a  body  falls  16  ft.  the  first  second,  48  ft.  the  next,  80 
the  next,  and  so  on,  constantly  increasing,  how  far  will  it  have  fallen  in 
4  sec. ;  in  4J  sec. ;  in  5  sec?  256  ft. ;  324  ft. ;  400  ft. 

95.  A  man  traveling  at  a  constant  increase,  is  observed  to  have  gone 
1  mile  the  first  hour,  3  miles  the  next,  5  the  next,  and  so  on :  how  far 
will  he  have  gone  in  6J  hr.?  42 J  mi. 

96.  The  number  of  men  in  a  side  rank  of  a  solid  body  of  militia,  is 
to  the  number  in  front  as  2  to  3;  if  the  length  and  breadth  be  in- 
creased so  as  to  number  each  4  men  more,  the  whole  body  will  contain 
2320  men :  how  many  does  it  now  contain?  1944  men. 

97.  A  grocer  at  one  straight  cut  took  off  a  segment  of  a  cheese 
which  had  J  of  the  circumference,  and  weighed  3  Ib. :  what  did  the 
whole  cheese  weigh?  33.0232+  Ib. 

98.  A  wooden  wheel  of  uniform  thickness,  4  ft.  in  diameter,  stands 
in  mud  1  ft.  deep:  what  fraction  of  the  wheel  is  out  of  the  mud? 

.80449  +  of  it. 

99.  My  lot  contains  135  sq.  rd.,  and  the  breadth  to  length  is  as  3  to 
5 :  what  is  the  width  of  a  road  which  shall  extend  from  one  corner 
half  round  the  lot,  and  occupy  J  of  the  ground  ?  24|  ft. 

100.  A  circular  lot  15  rd.  in  diameter  is  to  have  three  circular  grass 
beds  just  touching  each  other  and  the  large  boundary :  what  must  be 
the  distance  between  their  centers,  and  how  much  ground  is  left  in  the 
triangular  space  about  the  main  center? 

Distance,  6.9615242+  rd. 

Space  within,  1.9537115+  sq.  rd. 

101.  T  have  an  inch  board  5  ft.  long,  17  in.  wide  at  one  end,  and  7 
in.  at  the  other :  how  far  from  the  larger  end  must  it  be  cut  straight 
across,  so  that  the  solidities  of  the  two  parts  shall  be  equal?  2  ft. 

102.  Four  equal  circular  pieces  of  uniform  thickness,  the  largest 
possible,  are  to  be  cut  from  a  circular  plate  of  the  same  thickness,  and 
worth  $67 :  supposing  there  is  no  waste,  what  is  the  worth  of  each  of 
the  four,  and  what  is  the  worth  of  the  outer  portion  which  is  left? 

Each  small  circle,  $11.49538+ 
Outer  portion,        $17.87747+ 


408  RAY'S  HIGHER  ARITHMETIC. 

• 

103.  A  12-inch  ball  is  in  the  corner  where  walls  and  floor  are  at 
right  angles :  what  must  be  the  diameter  of  another  ball  which  can 
touch  that  ball  while  both  touch  the  same  floor  and  the  same  walls? 

3.2154  in.,  or  44.7846  in. 

104.  A  workman  had  a  squared  log  twice  as  long  as  wide  or  deep; 
lie  made  out  of  it  a  water-trough,  of  sides,  ends,  and  bottom  each  3 
inches  thick,  and  having  11772  solid  inches:  what  is  the  capacity  of  it 
in  gallons?  68T8T  gal. 

105.  How  many  inch  balls  can  be  put  in  a  box  which  measures, 
inside,  10  in.  square,  and  is  5  in.  deep?  568  balls. 

106.  A  tin  vessel,  having  a  circular  mouth  9  in.  in  diameter,  a 
bottom  4J  in.  in  diameter,  and  a  depth  of  10  in.,  is  J  part  full  of 
water:  what  is  the  diameter  of  a  ball  which  can  be  put  in  and  just 
be  covered  by  the  water?  6.1967  in. 


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